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Combining Functions

- Calculus
- Absolute Maxima and Minima
- Absolute and Conditional Convergence
- Accumulation Function
- Accumulation Problems
- Algebraic Functions
- Alternating Series
- Antiderivatives
- Application of Derivatives
- Approximating Areas
- Arc Length of a Curve
- Area Between Two Curves
- Arithmetic Series
- Average Value of a Function
- Calculus of Parametric Curves
- Candidate Test
- Combining Differentiation Rules
- Combining Functions
- Continuity
- Continuity Over an Interval
- Convergence Tests
- Cost and Revenue
- Density and Center of Mass
- Derivative Functions
- Derivative of Exponential Function
- Derivative of Inverse Function
- Derivative of Logarithmic Functions
- Derivative of Trigonometric Functions
- Derivatives
- Derivatives and Continuity
- Derivatives and the Shape of a Graph
- Derivatives of Inverse Trigonometric Functions
- Derivatives of Polar Functions
- Derivatives of Sec, Csc and Cot
- Derivatives of Sin, Cos and Tan
- Determining Volumes by Slicing
- Direction Fields
- Disk Method
- Divergence Test
- Eliminating the Parameter
- Euler's Method
- Evaluating a Definite Integral
- Evaluation Theorem
- Exponential Functions
- Finding Limits
- Finding Limits of Specific Functions
- First Derivative Test
- Function Transformations
- General Solution of Differential Equation
- Geometric Series
- Growth Rate of Functions
- Higher-Order Derivatives
- Hydrostatic Pressure
- Hyperbolic Functions
- Implicit Differentiation Tangent Line
- Implicit Relations
- Improper Integrals
- Indefinite Integral
- Indeterminate Forms
- Initial Value Problem Differential Equations
- Integral Test
- Integrals of Exponential Functions
- Integrals of Motion
- Integrating Even and Odd Functions
- Integration Formula
- Integration Tables
- Integration Using Long Division
- Integration of Logarithmic Functions
- Integration using Inverse Trigonometric Functions
- Intermediate Value Theorem
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- Jump Discontinuity
- Lagrange Error Bound
- Limit Laws
- Limit of Vector Valued Function
- Limit of a Sequence
- Limits
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- Limits of a Function
- Linear Approximations and Differentials
- Linear Differential Equation
- Linear Functions
- Logarithmic Differentiation
- Logarithmic Functions
- Logistic Differential Equation
- Maclaurin Series
- Manipulating Functions
- Maxima and Minima
- Maxima and Minima Problems
- Mean Value Theorem for Integrals
- Models for Population Growth
- Motion Along a Line
- Motion in Space
- Natural Logarithmic Function
- Net Change Theorem
- Newton's Method
- Nonhomogeneous Differential Equation
- One-Sided Limits
- Optimization Problems
- P Series
- Particle Model Motion
- Particular Solutions to Differential Equations
- Polar Coordinates
- Polar Coordinates Functions
- Polar Curves
- Population Change
- Power Series
- Radius of Convergence
- Ratio Test
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- Riemann Sum
- Rolle's Theorem
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- Second Derivative Test
- Separable Equations
- Separation of Variables
- Simpson's Rule
- Solid of Revolution
- Solutions to Differential Equations
- Surface Area of Revolution
- Symmetry of Functions
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- Taylor Polynomials
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- Techniques of Integration
- The Fundamental Theorem of Calculus
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- The Power Rule
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- Theorems of Continuity
- Trigonometric Substitution
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- Decision Maths
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- 2 Dimensional Figures
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- Composition
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- Convexity in Polygons
- Coordinate Systems
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- Distance and Midpoints
- Equation of Circles
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- Figures
- Fundamentals of Geometry
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- HL ASA and AAS
- Identity Map
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- Law of Cosines
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- Plane Geometry
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- Triangle Inequalities
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- Using Similar Polygons
- Vector Addition
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- Volume of Cone
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- Volume of Solid
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- Volume of prisms
- Mechanics Maths
- Acceleration and Time
- Acceleration and Velocity
- Angular Speed
- Assumptions
- Calculus Kinematics
- Coefficient of Friction
- Connected Particles
- Conservation of Mechanical Energy
- Constant Acceleration
- Constant Acceleration Equations
- Converting Units
- Elastic Strings and Springs
- Force as a Vector
- Kinematics
- Newton's First Law
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- Power
- Projectiles
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- Resolving Forces
- Statics and Dynamics
- Tension in Strings
- Variable Acceleration
- Work Done by a Constant Force
- Probability and Statistics
- Bar Graphs
- Basic Probability
- Charts and Diagrams
- Conditional Probabilities
- Continuous and Discrete Data
- Frequency, Frequency Tables and Levels of Measurement
- Independent Events Probability
- Line Graphs
- Mean Median and Mode
- Mutually Exclusive Probabilities
- Probability Rules
- Probability of Combined Events
- Quartiles and Interquartile Range
- Systematic Listing
- Pure Maths
- ASA Theorem
- Absolute Value Equations and Inequalities
- Addition and Subtraction of Rational Expressions
- Addition, Subtraction, Multiplication and Division
- Algebra
- Algebraic Fractions
- Algebraic Notation
- Algebraic Representation
- Analyzing Graphs of Polynomials
- Angle Measure
- Angles
- Angles in Polygons
- Approximation and Estimation
- Area and Circumference of a Circle
- Area and Perimeter of Quadrilaterals
- Area of Triangles
- Argand Diagram
- Arithmetic Sequences
- Average Rate of Change
- Bijective Functions
- Binomial Expansion
- Binomial Theorem
- Chain Rule
- Circle Theorems
- Circles
- Circles Maths
- Combination of Functions
- Combinatorics
- Common Factors
- Common Multiples
- Completing the Square
- Completing the Squares
- Complex Numbers
- Composite Functions
- Composition of Functions
- Compound Interest
- Compound Units
- Conic Sections
- Construction and Loci
- Converting Metrics
- Convexity and Concavity
- Coordinate Geometry
- Coordinates in Four Quadrants
- Cubic Function Graph
- Cubic Polynomial Graphs
- Data transformations
- De Moivre's Theorem
- Deductive Reasoning
- Definite Integrals
- Deriving Equations
- Determinant of Inverse Matrix
- Determinants
- Differential Equations
- Differentiation
- Differentiation Rules
- Differentiation from First Principles
- Differentiation of Hyperbolic Functions
- Direct and Inverse proportions
- Disjoint and Overlapping Events
- Disproof by Counterexample
- Distance from a Point to a Line
- Divisibility Tests
- Double Angle and Half Angle Formulas
- Drawing Conclusions from Examples
- Ellipse
- Equation of Line in 3D
- Equation of a Perpendicular Bisector
- Equation of a circle
- Equations
- Equations and Identities
- Equations and Inequalities
- Estimation in Real Life
- Euclidean Algorithm
- Evaluating and Graphing Polynomials
- Even Functions
- Exponential Form of Complex Numbers
- Exponential Rules
- Exponentials and Logarithms
- Expression Math
- Expressions and Formulas
- Faces Edges and Vertices
- Factorials
- Factoring Polynomials
- Factoring Quadratic Equations
- Factorising expressions
- Factors
- Finding Maxima and Minima Using Derivatives
- Finding Rational Zeros
- Finding the Area
- Forms of Quadratic Functions
- Fractional Powers
- Fractional Ratio
- Fractions
- Fractions and Decimals
- Fractions and Factors
- Fractions in Expressions and Equations
- Fractions, Decimals and Percentages
- Function Basics
- Functional Analysis
- Functions
- Fundamental Counting Principle
- Fundamental Theorem of Algebra
- Generating Terms of a Sequence
- Geometric Sequence
- Gradient and Intercept
- Graphical Representation
- Graphing Rational Functions
- Graphing Trigonometric Functions
- Graphs
- Graphs and Differentiation
- Graphs of Common Functions
- Graphs of Exponents and Logarithms
- Graphs of Trigonometric Functions
- Greatest Common Divisor
- Growth and Decay
- Growth of Functions
- Highest Common Factor
- Hyperbolas
- Imaginary Unit and Polar Bijection
- Implicit differentiation
- Inductive Reasoning
- Inequalities Maths
- Infinite geometric series
- Injective functions
- Instantaneous Rate of Change
- Integers
- Integrating Polynomials
- Integrating Trigonometric Functions
- Integrating e^x and 1/x
- Integration
- Integration Using Partial Fractions
- Integration by Parts
- Integration by Substitution
- Integration of Hyperbolic Functions
- Interest
- Inverse Hyperbolic Functions
- Inverse Matrices
- Inverse and Joint Variation
- Inverse functions
- Iterative Methods
- L'Hopital's Rule
- Law of Cosines in Algebra
- Law of Sines in Algebra
- Laws of Logs
- Limits of Accuracy
- Linear Expressions
- Linear Systems
- Linear Transformations of Matrices
- Location of Roots
- Logarithm Base
- Logic
- Lower and Upper Bounds
- Lowest Common Denominator
- Lowest Common Multiple
- Math formula
- Matrices
- Matrix Addition and Subtraction
- Matrix Determinant
- Matrix Multiplication
- Metric and Imperial Units
- Misleading Graphs
- Mixed Expressions
- Modulus Functions
- Modulus and Phase
- Multiples of Pi
- Multiplication and Division of Fractions
- Multiplicative Relationship
- Multiplying and Dividing Rational Expressions
- Natural Logarithm
- Natural Numbers
- Notation
- Number
- Number Line
- Number Systems
- Numerical Methods
- Odd functions
- Open Sentences and Identities
- Operation with Complex Numbers
- Operations with Decimals
- Operations with Matrices
- Operations with Polynomials
- Order of Operations
- Parabola
- Parallel Lines
- Parametric Differentiation
- Parametric Equations
- Parametric Integration
- Partial Fractions
- Pascal's Triangle
- Percentage
- Percentage Increase and Decrease
- Percentage as fraction or decimals
- Perimeter of a Triangle
- Permutations and Combinations
- Perpendicular Lines
- Points Lines and Planes
- Polynomial Graphs
- Polynomials
- Powers Roots And Radicals
- Powers and Exponents
- Powers and Roots
- Prime Factorization
- Prime Numbers
- Problem-solving Models and Strategies
- Product Rule
- Proof
- Proof and Mathematical Induction
- Proof by Contradiction
- Proof by Deduction
- Proof by Exhaustion
- Proof by Induction
- Properties of Exponents
- Proportion
- Proving an Identity
- Pythagorean Identities
- Quadratic Equations
- Quadratic Function Graphs
- Quadratic Graphs
- Quadratic functions
- Quadrilaterals
- Quotient Rule
- Radians
- Radical Functions
- Rates of Change
- Ratio
- Ratio Fractions
- Rational Exponents
- Rational Expressions
- Rational Functions
- Rational Numbers and Fractions
- Ratios as Fractions
- Real Numbers
- Reciprocal Graphs
- Recurrence Relation
- Recursion and Special Sequences
- Remainder and Factor Theorems
- Representation of Complex Numbers
- Rewriting Formulas and Equations
- Roots of Complex Numbers
- Roots of Polynomials
- Roots of Unity
- Rounding
- SAS Theorem
- SSS Theorem
- Scalar Triple Product
- Scale Drawings and Maps
- Scale Factors
- Scientific Notation
- Second Order Recurrence Relation
- Sector of a Circle
- Segment of a Circle
- Sequences
- Sequences and Series
- Series Maths
- Sets Math
- Similar Triangles
- Similar and Congruent Shapes
- Simple Interest
- Simplifying Fractions
- Simplifying Radicals
- Simultaneous Equations
- Sine and Cosine Rules
- Small Angle Approximation
- Solving Linear Equations
- Solving Linear Systems
- Solving Quadratic Equations
- Solving Radical Inequalities
- Solving Rational Equations
- Solving Simultaneous Equations Using Matrices
- Solving Systems of Inequalities
- Solving Trigonometric Equations
- Solving and Graphing Quadratic Equations
- Solving and Graphing Quadratic Inequalities
- Special Products
- Standard Form
- Standard Integrals
- Standard Unit
- Straight Line Graphs
- Substraction and addition of fractions
- Sum and Difference of Angles Formulas
- Sum of Natural Numbers
- Surds
- Surjective functions
- Tables and Graphs
- Tangent of a Circle
- The Quadratic Formula and the Discriminant
- Transformations
- Transformations of Graphs
- Translations of Trigonometric Functions
- Triangle Rules
- Triangle trigonometry
- Trigonometric Functions
- Trigonometric Functions of General Angles
- Trigonometric Identities
- Trigonometric Ratios
- Trigonometry
- Turning Points
- Types of Functions
- Types of Numbers
- Types of Triangles
- Unit Circle
- Units
- Variables in Algebra
- Vectors
- Verifying Trigonometric Identities
- Writing Equations
- Writing Linear Equations
- Statistics
- Bias in Experiments
- Binomial Distribution
- Binomial Hypothesis Test
- Bivariate Data
- Box Plots
- Categorical Data
- Categorical Variables
- Central Limit Theorem
- Chi Square Test for Goodness of Fit
- Chi Square Test for Homogeneity
- Chi Square Test for Independence
- Chi-Square Distribution
- Combining Random Variables
- Comparing Data
- Comparing Two Means Hypothesis Testing
- Conditional Probability
- Conducting a Study
- Conducting a Survey
- Conducting an Experiment
- Confidence Interval for Population Mean
- Confidence Interval for Population Proportion
- Confidence Interval for Slope of Regression Line
- Confidence Interval for the Difference of Two Means
- Confidence Intervals
- Correlation Math
- Cumulative Distribution Function
- Cumulative Frequency
- Data Analysis
- Data Interpretation
- Degrees of Freedom
- Discrete Random Variable
- Distributions
- Dot Plot
- Empirical Rule
- Errors in Hypothesis Testing
- Estimator Bias
- Events (Probability)
- Frequency Polygons
- Generalization and Conclusions
- Geometric Distribution
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- Hypothesis Test for Correlation
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- Hypothesis Test of Two Population Proportions
- Hypothesis Testing
- Inference for Distributions of Categorical Data
- Inferences in Statistics
- Large Data Set
- Least Squares Linear Regression
- Linear Interpolation
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- Measures of Central Tendency
- Methods of Data Collection
- Normal Distribution
- Normal Distribution Hypothesis Test
- Normal Distribution Percentile
- Paired T-Test
- Point Estimation
- Probability
- Probability Calculations
- Probability Density Function
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- Quantitative Variables
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- Sample Mean
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- Variance for Binomial Distribution
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Jetzt kostenlos anmelden**Combining functions **is one way to manipulate functions in AP Calculus. It allows us to create new functions from existing ones. While most of combining functions are the result of algebraic operations (think addition/subtraction and multiplication/division), another way of combining functions is a new method we will discuss called **composition of functions**.

In this article, we cover all things related to the combining and composing of functions.

- Arithmetic combinations of functions
- Function composition
- Combining piecewise functions
- Combining functions: examples and real-world applications

We can create a new function by adding, subtracting, multiplying, or dividing functions, just like how these operations form a new number.

The **combination of functions** is the act of combining multiple functions into a single function. This usually entails using basic **mathematical operators **such as addition or multiplication.

To start, let's take a look at the easier way to combine functions: **arithmetic combination**** of functions**.

**Arithmetic combination of functions** is the creation of new functions by combining existing functions using basic arithmetic operations such as: addition, subtraction, multiplication, and division.

Just as numbers can be added, subtracted, multiplied, and divided, we can use these same operations on functions!

Consider the functions \(f(x) = 2x + 1\) and \(g(x) = 3x +2\).

What happens when we combine them by addition?

\[(f+g)(x) = f(x) + g(x)\]

Now we can substitute each of the functions.

\[(f + g)(x) = (2x +1) + (3x +2)\]

And finally, collect like terms.

\[(f + g)(x) = 5x + 3\]

Let's consider the same functions again, \(f(x) = 2x + 1\) and \(g(x) = 3x +2\).

What happens when we combine them by subtraction?

\[(f-g)(x) = f(x) - g(x)\]

Now we substitute the functions as before.

\[(f - g)(x) = (2x +1) - (3x +2)\]

And we can collect like terms.

\[(f - g)(x) = -x - 1\]

Let's consider the functions \(f(x) = 2x + 4\) and \(g(x) = x + 2\)

What happens if we multiply them together.

\[(fg)(x) = f(x) \cdot g(x)\]

And substitute in the functions.

\[(fg)(x) = (2x + 4)(x+2)\]

We then multiply out the brackets.

\[(fg)(x) = 2x^2 + 4x + 4x + 8\]

And again we can collect like terms.

\[(fg)(x) = 2x^2 + 8x + 8\]

Let's consider the functions \(f(x) = 2x + 4\) and \(g(x) = x + 2\) once again.

What happens when we combine them by division.

\[(\frac{f}{g})(x) = \frac{f(x)}{g(x)}\]

And substitute in the functions.

\[(\frac{f}{g})(x) = \frac{2x+4}{x+2}\]

Then we simplify algebraically. In the case of these functions, we find a common factor and divide through.

\[(\frac{f}{g})(x) = \frac{2(x+2)}{x+2}\]

\[(\frac{f}{g})(x) = 2\]

In most precalculus maths, the main object we study is numbers. While we also study overall patterns as seen in functions and equations, this usually occurs through examining the numbers themselves.

This changes in calculus. In calculus, the fundamental objects being studied are the functions themselves, which are much more sophisticated mathematical objects than numbers.

It is often helpful to look at a function's formula and observe its algebraic structure. For instance, given the quadratic function,

\[f(x) = -3x^2 + 5x -7\]

we might benefit from thinking of this as the sum of three simpler functions:

- the constant function \(c(x) = -7\),
- the linear function \(l(x) = 5x\) that passes through the point (0, 0) with a slope of m = 5, and
- the concave down quadratic function \(q(x) = -3x^2\).

Each of the simpler functions: c, l, and q contribute to making f the function that it is.

Likewise, if we were interested in the function,

\[p(x) = (3x^2 + 4)(9-2x^2)\]

it might be natural for us to think about the two simpler functions:

- \(f(x) = 3x^2 + 4\)and
- \(g(x) = 9 -2x^2\)

that are multiplied together to produce p.

When we arithmetically combine functions, the **domain **of the new function will contain the x-values that are common between the original functions. In other words, both functions must be defined at a point for their combination to be defined.

Additionally, when dividing functions, the domain is further restricted so that the denominator isn't equal to zero.

Basically, what this means is when we evaluate combined functions, we can either:

**Combine**the functions and**then evaluate**, or- We can
**evaluate**the functions and**then combine**them.

If the domain of a function, f, is the set, A, and the domain of the function, g, is the set, B, then the domain of f + g is the intersection \(A \cap B\) (note that the symbol, \(\cap\), just means "intersection") because both f(x) and g(x) have to be defined.

Let's say we have the functions:

\[f(x) = \sqrt x g(x) = \sqrt{2-x}\]

The domain of f(x) is \(A = [0, \infty )\).

The domain of g(x) is \(B = (-\infty, 2]\).

So, the domain of (f + g)(x) is:

\((f + g) (x) = \sqrt x + \sqrt{2-x}, \text{ Domain of } (f + g)(x) \text{ is }A \cap B = [0, 2]\)

Similarly, the domain of (f - g)(x) is:

\((f - g) (x) = \sqrt x - \sqrt{2-x}, \text{ Domain of } (f - g)(x) \text{ is }A \cap B = [0, 2]\)

As with the addition and subtraction of functions, the domain for the multiplication and division of functions is the intersection \(A \cap B\).

However, when we divide functions, we need to further restrict the domain of the combined function since we can't divide by zero. So, for the division of functions, the domain is \(x \in A \cap B \space | \space g(x) ≠ 0\) . This is read as "the set of all values of x such that x is an element of the intersection of A and B, as long as g(x) does not equal zero." (That is quite the mouthful!)

Let's say we have the functions:

\[f(x) = x^2; \space g(x) = x-1\]

The domain of f(x) is \(A = (-\infty, \infty)\).

The domain of g(x) is \(B = (-\infty, \infty)\).

So, the domain of \((f \cdot g) (x)\) is:

\((fg)(x) = x^2(x-1); \text{ Domain of } (fg)(x) \text{ is } A\cap B = (-\infty, \infty)\)

But, the domain of \((\frac{f}{g})(x)\) is:

\((\frac{f}{g})(x) = \frac{x^2}{x-1}; \text{ Domain of } (\frac{f}{g})(x) \text{ is } A \cap B= (-\infty, 1)\cup (1, \infty)\)

If we consider two functions: f(x) and g(x), then, for the values of x within the domain of both f(x) and g(x), the sum, difference, product, and quotient of the two functions are defined as:

- Sum: \((f + g) (x) = f(x) + g(x)\)
- Difference: \((f - g) (x) = f(x) - g(x)\)
- Product: \((f \cdot g) (x) = f(x) \cdot g(x)\)
- Quotient: \((\frac{f}{g}) (x) = \frac{f(x)}{g(x)}; g≠0\)

For these next four sections, let's consider the two functions:

\(f(x) = 5x + 2\) and \(g(x) = x^2-1\)

We will also solve each combination where x = 4.

\[f(4) = 5(4) + 2 = 20 + 2 = 22\]

\[g(4) = (4)^2 - 1 = 16-1 = 15\]

What is \((f+g)(x)\)?

Operation | Combine, then Evaluate | Evaluate, then Combine | Domain | ||||

\((f+g)(x)\) | \((f+g)(x)= f(x) +g(x)\) | \(= (5x+2) + (x^2 -1)\) | \(= 5x + 2+x^2 -1\) | \(=x^2+5x+1\) | --- | --- | \((-\infty, \infty)\) |

\((f+g)(4)= f(4) +g(4)\) | \(= (5(4)+2) + ((4)^2 -1)\) | \(= 5(4) + 2+(4)^2 -1\) | \(16 + 20 +1\) | \((f+g)(4)= f(4) +g(4)\) | \(=22+15 = 37\) |

What is \((f-g)(x)\)?

Operation | Combine, then Evaluate | Evaluate, then Combine | Domain | ||||

\((f-g)(x)\) | \((f-g)(x) = f(x)-g(x)\) | \(= (5x+2) - (x^2 -1)\) | \(=5x + 2 - x^2 + 1\) | \(-x^2+5x+3\) | --- | --- | \((-\infty, \infty)\) |

\((f-g)(4) = f(4)-g(4)\) | \(= -(4)^2 + 5(4) + 3\) | \(=-16 + 20 + 3\) | \(= 7\) | \(f(4) - g(4)\) | \(22-15 = 7\) |

What is \((f \cdot g)(x)\)?

Operation | Combine, then Evaluate | Evaluate, then Combine | Domain | ||||

\((f \cdot g)(x)\) | \((f \cdot g)(x) = f(x) \cdot g(x)\) | \(= (5x+2) (x^2 -1)\) | \(=5x^3-5x+2x^2-2\) | \(=5x^3+2x^2-5x-2\) | --- | --- | \((-\infty, \infty)\) |

\((f \cdot g)(4) = f(4) \cdot g(4)\) | \(= 5(4)^3 + 2(4)^2-5(4)-2\) | \(320 +32-20-2\) | \(=330\) | \((f \cdot g)(4)\) | \(22 \cdot 15 = 330\) |

What is \((\frac{f}{g})(x)\)?

Operation | Combine, then Evaluate | Evaluate, then Combine | Domain | |||

\((\frac{f}{g})(x)\) | \((\frac{f}{g})(x) = \frac{5x+2}{x^2-1}\) | --- | --- | --- | --- | \((\infty, -1) \cup (-1, 1) \cup (1, \infty)\) |

\((\frac{f}{g})(4)\) | \((\frac{f}{g})(4)= \frac{5(4)+2}{(4)^2-1}\) | \(\frac{20+2}{16-1}= \frac{22}{15}\) | \(\frac{f(4)}{g(4)}\) | \(\frac{f(4)}{g(4)} = \frac{22}{15}\) |

When we divide functions, the domain is restricted so that the denominator does not equal zero.

Functions can also be combined by a process called **function composition** (or, the composition of functions, both mean the exact same thing), which is when one function is composed with another by plugging one function into the other and solving.

For instance, say we have the functions:

\[y = f(u) = \sqrt{u} \text{ and } u=g(x)=x^2+1\]

Since y is a function of u and u is a function of x, it follows that y is ultimately a function of x. We can compute this by **substitution**:

\[y = f(u)= f(g(x)) = f(x^2 +1) = \sqrt{x^2 +1}\]

This process is called **composition **because the new function is **composed **of the two original functions, f and g.

In general, if we want to compose any two functions, f and g:

- We start with a number x in the domain of gand find g(x).
- If g(x) is in the domain of f, then
- We can calculate the value of f(g(x)).

Note that the output of the first function (g(x) in this case) is used as the input to the next function (f(x) in this case).

The result of this composition is the new function \(h(x) = f(g(x))\) which is obtained by substituting g into f. This is called the composition (or composite) of f and g and is denoted by \((f \circ g)\) read as "fog" or "f circle g" or "f of g of x".

**Function composition** involves taking one function, say \(g(x)\), and plugging it into another function, say \(f(x)\), and either simplifying or solving for a value of x.

The composite function \((f \circ g)\) is defined as:

\((f \circ g)(x) = f(g(x))\)

An important note here: function composition is NOT multiplication!

The image below shows us how to picture \((f \circ g)\) in terms of machines.

Let's say we have two functions:

\(f(x) = 2 + 3x - x^2\) and \(g(x) = 2x -1\)

What is \((f \circ g)(x)\)?

Solution:

- The first step is to understand what \((f \circ g)(x)\) is telling us to do. Written in this order, we need to plug the function g(x) into the function f(x) and solve.
- From there, we need to remember the formula and write it down. So,
- \((f \circ g)(x) = f(g(x))\)

- Since we know the formula for g(x), let's plug it in:
- \(f(g(x)) = f(2x-1)\)

- Now, everywhere in the function f, replace x with 2x - 1:
- \((f \circ g)(x) = f(2x-1) = 2+3(2x-1)-(2x-1)^2\)

- And simplify:
- \((f \circ g)(x) = 2 + 3(2x-1) - (2x-1)^2 \rightarrow (f \circ g)(x) = 2 + 6x-3-(4x^2-4x+1)\rightarrow (f\circ g)(x) = -1 + 6x-4x^2 + 4x-1 \rightarrow (f \circ g)(x) = -4x^2 + 10x -2\)

It is important to note that when combining functions, the order of combination matters. In other words, **f(g(x))** **does not necessarily equal** **g(f(x))**. In fact, it is a special case when these two are equal!

In the example above, we had two functions and combined them using function composition. What if we reversed the order of the composition of these two functions? Let's take a look!

Let's take the same two functions in the previous example: \(f(x) = 2 + 3x - x^2\) and \(g(x) = 2x -1\)

Only this time, let's find \((g \circ f)(x)\).

Solution:

- Again, we first need to understand what \((g \circ f)(x)\) is telling us to do. Written in this order, we need to plug the function f(x) into the function g(x) and solve.
- So let's write the formula:
- \((g \circ f)(x) = g(f(x))\)

- Since we know the formula for f(x), let's plug it in:
- \(g(f(x)) = g(2 + 3x - x^2)\)

- Now, everywhere in the function g, replace x with \(2+3x-x^2\):
- \((g \circ f)(x) = g(2+3x-x^2) = 2(2 + 3x - x^2) - 1\)

- And simplify:
- \((g \circ f)(x) = 2(2 + 3x - x^2) - 1 \rightarrow (g \circ f)(x) = 4 + 6x - 2x^2 -1 \rightarrow (g \circ f)(x) = -2x^2 + 6x + 3\)

As we can see, this solution is **not **the same as the solution from the previous example. When **composing functions**, the **order matters**!

It is entirely possible to compose a function with itself!

This looks like: \((f \circ f)(x) = f(f(x))\)

So, if we have a function:

\[f(x) = 2x +3\]

and compose it with itself, we get:

\((f \circ f)(x) = f(f(x)) = f(2x + 3) = 2(2x+3)+3 = 4x + 6 +3 = 4x + 9\)

Don't forget! We need to consider **two domains** when composing functions. If we are trying to evaluate \((f \circ g)(x)\), we can see that g is evaluated at x, so xmust be in the domain of g. And since fis evaluated at g, g must be in the domain of f. In other words:

- For \((f \circ g)(x)\), x must be a value that can be plugged into g to give us a value of g(x) that can be plugged into f to get f(g(x)).
- For \((g \circ f)(x)\), x must be a value that can be plugged into f to give us a value of f(x) that can be plugged into g to get g(f(x)).

Let's consider the functions: \(f(x) = \sqrt{x}\) and \(g(x) = x^2\)

The domain of \(f(x) = \sqrt{x}\) is \([0, \text{ is } \infty)\).

The domain of \(g(x) = x^2\) is \((-\infty, \infty)\)

If we compose them as:

\((f \circ g)(x) = f(g(x))\)

we get:

\(f(g(x)) = f(x^2) = \sqrt{x^2} = x\)

- In this case, since both x and g(x) have a domain of all real numbers:
- The domain is all real numbers: \((-\infty, \infty)\)

**But**, if we compose them as:

\[(g \circ f)(x) = g(f(x))\]

we get:

\[g(f(x)) = g(\sqrt x) = (\sqrt{x})^2 = x\]

- In this case, while x would normally have a domain of all real numbers, we must consider f(x), which has a domain of all non-negative real numbers, so:
- The domain is all non-negative real numbers: \([0, \infty)\)

With all this talk about composing functions, it is worth mentioning that we can go the other way and **decompose functions** as well. This can be helpful if we are working on a more complicated function. Function decomposition is the term for breaking up a function into multiple, simpler functions.

Much like *there's no wrong way to eat a Reese's*, there is often more than one way to decompose a function!

Let's write the function:

\[f(x) = \sqrt{5 -x^2}\]

as the composition of two simpler functions.

Solution 1:

We are trying to break up, or **decompose,** f(x) into two simpler functions that we will call g(x) and h(x). So,

\[f(x) = g(h(x))\]

- To break up the function, we first must look for a
**function inside a function**within f(x).- As one option, we might see that the
**inside function**is \(5-x^2\), and the**outside function**is \(\sqrt x\).

- As one option, we might see that the
- After observing this, we could decompose the function as
- \(h(x) = 5 -x^2; \space g(x) = \sqrt x\)

- The final step is to double-check our answer by recomposing the functions:
- \(g(h(x)) = g(5 -x^2) = \sqrt{5-x^2} = f(x)\)

Solution 2:

- Another option would be to break up f(x) into g(x) and h(x) as:
- \(h(x) = x^2g(x) = \sqrt{5-x}\)

- Again, we need to double-check our answer by recomposing the functions:
- \(g(h(x)) = g(x^2) = \sqrt{5-x^2} = f(x)\)

Function composition is not the same as multiplying functions

Despite what the notation looks like, when we compose functions, we are not simply multiplying them together. When we compose functions, we are replacing the variable of the "outer" function with the "inner" function and evaluating.

Say we want to compose two functions:

\[f(x) = \sqrt x; \space g(x) = x-1\]

If we want to compose them as \((f \circ g)(x)\):

- Write the formula - \((f \circ g)(x) = f(g(x))\)
- This makes it more clear that
- f(x) is the "outer" function and
- g(x) is the "inner" function

- This makes it more clear that
- Now, replace the variable, x, of the "outer" function, f(x), with the "inner" function, g(x) to get:
- \((f \circ g)(x) = f(g(x)) = f(x-1) = \sqrt{x-1}\)
- So, the composed function is: \(f(g(x)) = \sqrt{x-1}\)

If we compare the composition, \((f \circ g)(x)\), to the multiplication, \((f \cdot g)(x)\), (this could also be written as (fg)(x) we see that the answers are **not the same**.

- When we multiply \((f \cdot g)(x)\), we get:
- \((f \cdot g)(x) = \sqrt x \cdot (x-1) = (\sqrt x)(x) - \sqrt x = x \sqrt x - \sqrt x\)
- Which is not the same as the composition. So, \((f \circ g) (x) ≠ (f \cdot g)(x)\)!

Unlike combining functions, **function composition**** is not commutative**.

\[(f \circ g)(x) ≠ (g \circ f) (x)\]

Now, it must be said that there are some **s****pecial cases** where \((f \circ g)(x)\) **does equal** \((g \circ f)(x)\). If this is the case, that means the two functions are **inverse functions**.

Function composition is, however, associative.

If we have three functions that can be composed with each other, the associative property applies.

Given f, g, and h: \((f \circ (g \circ h))(x) = f((g \circ h)(x)) = f(g(h(x))) = (f \circ g)(h(x)) = ((f \circ g) \circ h)(x)\)

So, \(f \circ(g \circ h) = (f \circ g) \circ h\)

When combining piecewise functions, we follow all the same rules as when combining any other functions, except:

- We might need to split the combination into multiple cases if only one function is defined at a point.

Say we have two piecewise functions:

\(f(x) = \begin{cases} 2 & x > 2 \\ 2 & x < 2 \end{cases}\) and \(g(x) = \begin{cases} -2 & x \geq 2 \\ 2 & x<2 \end{cases}\)

And we want to calculate g(x) - f(x).

Solution:

For this piecewise function, we need to split it into cases.

- For \(x <2\):
- f(x) = 3 and g(x) = 2.
- So, \((g - f)(x) = -1\)

- f(x) = 3 and g(x) = 2.
- At \(x = 2\):
- f(x) doesn't exist
- So, \((g - f)(x)\) is undefined

- f(x) doesn't exist
- For \(x > 2\):
- \(f(x) = 2\) and \(g(x) = -2\).
- So, \((g - f)(x) = -4\)

- \(f(x) = 2\) and \(g(x) = -2\).

Here are some more examples of combining functions!

If \(f(x) = x^2 - 2x - 3\) and \(g(x) = x + 1\), find the following:

- \((f-g)(-1)\)
- \(\left( \frac{f}{g} \right) (x)\)

Solutions:

- Here we need to subtract g(x) from f(x) and then substitute -1 for x.
- \((f-g)(x) = f(x) - g(x) = (x^2 -2x -3)-(x+1) = x^2 -2x - 3 -x -1 = x^2 - 3x -4\)
- \((f -g)(-1) = (-1)^2 -3(-1) -4 = 1 + 3 -4 = 0\)

- Here we need to divide f(x) by g(x). Note that we can factor f, so the fraction can be simplified.
- \(\left( \frac{f}{g} \right) = \frac {x^2-2x-3}{x+1} = \frac {(x-3)\cancel{(x+1)}}{\cancel{x+1}} = x-3 \text{ if } x≠ 1\)

Find \((f+g), (f-g), (fg),\text{ and } \left( \frac{f}{g} \right)\) for the functions:

\[f(x) = x^2 + 3g(x) = x-1\]

Solutions:

- \((f + g) = f(x) + g(x) = (x^2 + 3) + (x-1) = x^2 + 3 +x -1 = x^2 + x +2\)
- \((f-g) = f(x) - g(x) = (x^2+3) - (x-1) = x^2 + 3 - x +1 = x^2 - x + 4\)
- \((fg) = f(x) \cdot g(x) = (x^2 + 3)(x-1) = x^3 - x^2+3x-3\)
- \(\left( \frac{f}{g} \right) = \frac{f(x)}{g(x)} = \frac{x^2+3}{x-1}, x ≠ 1\)

The graph of the function f is shown in blue, and the graph of the function g is shown in green. Use the graphs to solve each quantity.

- \(f(3)\)
- \(g(3)\)
- \(f(3) + g(3)\)
- \((f -g) (3)\)

Solutions:

- \(f(3) = -3\)
- \(g(3) = 9\)
- \(f(3) + g(3) = -3 + 9 = 6\)
- \((f -g) (3) = f(3) - g(3) = -3 -9 = -12\)

If \(f(x) = x^2 + 10\) and \(g(x) = \sqrt{x-1}\), find the following:

- \(f(g(x))\)
- \((g \circ f)(4)\)
- \((f\circ f)(x)\)

Solutions:

- Here we need to substitute g(x) for the x in f(x).
- \(f(g(x)) = f(\sqrt{x-1}) = (\sqrt{x-1})^2 +10\)
- \(f(g(x)) = x-1+10\)

- Here we need to find g(f(x)) and then evaluate at x = 4.
- \((g \circ f)(x) = g(f(x)) = g(x^2 + 10) = \sqrt{(x^2 + 10) - 1} = \sqrt{x^2 + 9}\)
- \((g \circ f)(4) = \sqrt{(4)^2 + 9} = \sqrt{16+9} = \sqrt{25} = 5\)

- Here we are composing f with itself, so we substitute \(x^2-10\) for the x in f.
- \((f \circ f)(x) = f(f(x)) = f(x^2 + 10) = (x^2 +10)^2 + 10\)
- \((f \circ f)(x) = x^4 + 20 x^2 + 110\)

If \(f(x) = \sqrt x\) and \(g (x) = \sqrt{2-x}\), find each composition and its domain.

- \((f \circ g)\)
- \((g \circ f)\)
- \((f \circ f)\)
- \((g \circ g)\)

Solutions:

- \(f \circ g = f(g(x))\)
- \(f(g(x)) = f(\sqrt{2-x}) = \sqrt {\sqrt{2-x}} = \sqrt[4]{2-x}\)
- The domain of \(f(g(x))\) is \((-\infty, 2]\)

- \(g \circ f = g(f(x))\)
- \(g(f(x))= g(\sqrt{x}) = \sqrt{2-\sqrt{x}}\)
- To find the domain, both \(\sqrt x\) and \(\sqrt{2-\sqrt{x}}\) must have \(x \geq 0\).
- For \(\sqrt{x}\) to be defined, \(x \geq 0\).
- For \(\sqrt{2-\sqrt x}\) to be defined, \(2-\sqrt x \geq 0\), or \(\sqrt x \leq 2\), or \(x \leq 4\).
- So, to meet both requirements, the domain of \(g(f(x))\) is \([0, 4]\)

- \(f \circ f = f(f(x))\)
- \(f(f(x)) = f(\sqrt{x}) = \sqrt {\sqrt{x}} = \sqrt[4]{x}\)
- The domain of \(f(f(x))\) is \([0, \infty))

- \(g \circ g = g(g(x))\)
- \(g(g(x)) = g(\sqrt{2-x}) = \sqrt{2 - \sqrt{2-x}}\)
- To find the domain, both \(2 - x \leq 0\) and \(2 - \sqrt{2-x} \geq 0\).
- For \(2-x \geq 0, \space x \leq 2\)
- For \(2 - \sqrt{2-x} \geq 0, \space \sqrt{2-x} \leq 2 \rightarrow 2 - x \leq 4 \rightarrow x \geq -2\)
- So, the domain of \(g(g(x))\) is \([-2, 2]\)

We can also compose three or more functions! For instance, the composite function \(f \circ g \circ h\) can be found by first applying h, then g, and then f like so:

\[(f \circ g\circ h)(x) = f(g(h(x)))\]

So, find \((f \circ g \circ h)(x)\) if:

\[f(x) = \frac{x}{x+1} \qquad g(x) = x^{10} \qquad h(x) = x +3 \]

Solution:

\[(f \circ g \circ h)(x) = f(g(h(x))) = f(g(x+3)) = f \left( (x+3)^{10}\right) = \frac{(x+3)^{10}}{(x+3)^{10}+1}\]

Given the function:

\[F(x) = \cos^2(x+9)\]

Find functions f, g, and h such that \(F = f \circ g \circ h\).

Solution:

- Since \(F(x) = (\cos(x+9))^2\), the formula for F says:
- First add 9
- Then take the cosine of the result
- And square it.

- So, we let:
- \(h(x) = x +9\)
- \(g(x) = \cos(x)\)
- \(f(x) = x^2\)

- Therefore, \((f \circ g \circ h)(x) = f(g(h(x))) = f(g(x+9)) = f(\cos(x+9)) = (\cos(x+9))^2 = F(x)\)

Graphing combinations of functions can be done simply in two steps:

Combine the relevant functions into a single function.

Graph the newly created function.

What is really interesting, however, is seeing how combining two functions will alter the original graphs of those functions. For instance, if we combined the functions \(f(x) = 3x + 4\) and \(g(x) = x\), how do you think the graph of our combined function would compare to the graph of f(x)? Let's work through it and find out.

First, let's simply combine the functions f(x) and g(x) into a single function h(x) by addition.

\[h(x) = (f+g)(x) = f(x) + g(x)\]

And next as always we substitute in our functions f(x) and g(x).

\[h(x) = 3x + 4+ x\]

And as before, we simply gather like terms to find our final combined function.

\[h(x) = 4x +4\]

Plotting the two original functions on a graph with the new combined function gives us the plot below. What can we notice about how h(x) relates to f(x) and g(x)? Well, as we can see, h(x) has a greater gradient than either f(x) or g(x). In fact, the gradient of h(x) is equal to the gradients of f(x) and g(x) combined!

When we consider what we are really doing in combining functions, this makes perfect sense. The output of a combined function (the number on the **vertical** axis) for a given input (the number on the **horizontal** axis) is simply a combination of the outputs of each original function.

To test this, pick a value on the *x-axis, *and find the *y-axis *value for each of the functions. Does the *y-axis* value for h(x) equal the *y-axis *values for f(x) and g(x) added together?

Combination of two functions on a graph.

- There are two main ways to combine functions:
- Combine functions using algebraic operations (a.k.a. arithmetic combinations of functions)
- Function composition

- Arithmetic combinations of functions uses precalculus methods to combine existing functions to create new ones:
- Addition and subtraction
- Multiplication and division

- Function composition uses a method of substitution to create new functions from existing ones.
- When combining functions, the domain of the new one always needs to be considered.
- Common mistakes when combining functions are:
- Forgetting to consider the domain
- Treating function composition the same as multiplying functions
- Forgetting that the order of composition of functions matters

\((f+g)(x)= f(x) +g(x)\)g(g(x)) = g(\sqrt{2-x}) = \sqrt{2 - \sqrt{2-x}}\)

There are two main ways to combine functions:

- Arithmetic combination and
- Function composition.

Arithmetic combination is when you combine functions using arithmetic means:

- Addition & Subtraction
- Multiplication & Division

Function composition is when you plug one function into another function and evaluate.

These can be combined in the same manner as any other two functions:

- Arithmetic combination and
- Function composition

However, exponential and logarithmic functions are inverses of each other, meaning they can "undo" each other if they share the same base.

If you plug an exponential function into a logarithmic function of the same base, you get the same answer if you reversed the order.

Piecewise functions follow the same rules as when combining any other type of function, except:

- You might need to split the combination into cases if only one function is defined at a point.

Combination of functions is the creation of new functions by combining existing functions. This can be done by basic arithmetic:

- Addition
- Subtraction
- Multiplication
- Division

Functions can also be combined by a process called function composition, which is when one function is composed with another.

So, if we have two functions, f and g, and we want to compose them, we do so by plugging one function into the other and solving.

Let's say we want to compose f and g by plugging g into f.

- First, x is plugged into g and g is evaluated.
- Second, the result of that is then plugged into f and evaluated.

Let's say you buy something at a hardware store, but what you buy is too big to fit in your car. For a fee, you can have the hardware store deliver your purchase for you. You pay for your purchase, plus the sales taxes, plus the fee. Taxes are 7.5%, the fee is $20.

Write a function t(x) for the total, after taxes, on the purchase amount x.

Write another function f(x) for the total, including the delivery fee, on the purchase amount x.

Calculate and interpret (f ∘ t)(x) and (t ∘ f)(x). Which results in a lower cost to you?

- Suppose taxes, by law, are not to be charged on delivery fees. Which composite function must then be used?

This sort of calculation actually comes up in real life, and is used for programming the cash registers.

- The taxes are 7.5%, so the tax function is given by t(x) = 1.075x
- The delivery fee is fixed, so the purchase amount is irrelevant.
- The fee function is given by f(x) = x + 20

- Composing, we get:
- (f ∘ t)(x) = f(t(x)) = f(1.075x) = 1.075x + 20
- (t ∘ f)(x) = t(f (x)) = t(x + 20) = 1.075(x + 20) = 1.075x + 21.50
- So, you would pay more using (t ∘ f)(x), because you would be paying taxes (the t(x) formula) on the delivery fee (the "+20" in the f (x) formula).
- You want the delivery fee to be tacked on after the taxes, because (f ∘ t)(x) results in a lower cost.

- If the state is not allowed to collect taxes on delivery fees, then:
- The function to use is (f ∘ t)(x).

More about Combining Functions

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