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Growth of Functions

- Calculus
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Ask yourself this question: Would you rather have \($1,000\) now, with an extra \($100\) added to at the end of each year, or would you rather have that money in a bank account that pays \(6\%\) annual interest compounded semi-annually? At first glance, it might be difficult to work out which option will give you the most profit in the long term. Therefore, it might be a good idea to do some calculations and even draw a graph to help you make your decision, based on how much your money will grow, don't you think? Both of the options given above are examples of **growth**. However, they do so in different ways; linear growth and geometric growth, respectively.

In this article, we will define what growth is, and why is modelling growth so important. We will also explore the main types of models that we can use to model growth, and the formulas and calculations used to determine each type of growth, using practical examples.

Let's start with the basics and define what we mean by **growth**.

**Growth** is a value of increase or decrease in relation to a previous value or an initial value.

Modelling growth and familiarizing yourself with the different types of growth can be very useful in life. If we think about it for a moment, we are surrounded by growth in our daily lives. We see growth in economy, population, finances, knowledge, skills, nature, and many others areas. Some of these areas experiment a faster growth than others depending on many aspects that might affect them internally and externally.

For example, **the population of a country can grow faster or slower than others**, depending on factors like quality of health services, education, access to resources, income levels, birth rates, mortality rates, and migration, among others.

Modelling growth is important because it helps us understand and analyze growth, helps us to predict future values according to a current trend, and triggers decision-making when required. For example, decisions might need to be made and actions taken if the population of a country is growing too rapidly or too slowly.

Now that we have a general idea of what growth is, let's focus on the **growth of functions**, which is determined based on their** growth factor.**

The **growth factor** is a factor by which a quantity increases or decreases per unit of another quantity.

This factor **can be a constant or a rate**, depending on which type of growth we are dealing with.

Remember the question at the beginning of the article, in that case, the extra \($100\) added to at the end of each year is the growth factor in the first option (a fixed amount). For the second option, you can use the growth rate of \(6\%\) to calculate the growth factor. The second option grows in a different way, as it represents a percentage or rate of the current value, rather than a fixed amount.

We will explain these cases more in depth when we look into linear and geometric growth, later in the article.

I think you would agree that a quadratic function increases and decreases faster than in a linear function.

Growth of a linear function vs a quadratic function.

Why is that exactly? Let's look into the reasons behind this in the section below.

When we are **comparing the growth of polynomial functions** like the ones in the previous example, we need to **look at the term with the highest order** in the equation of each function and consider an **appropriate interval of values of **\(x\).

In the example above, the term with the highest order in \(f(x)=x\) is \(x\), with a degree of 1.

The term with the highest order in \(f(x)=x^2\) is \(x^2\), with a degree of 2.

We can see that \(2>1\), therefore, for \(x>0\), the quadratic function will grow faster than the linear function.

What happens when you have **polynomial**** ****functions of the same order**? How do you compare their growth?

If you have polynomial functions of the same order such as \(f(x)=x^2+3x\) and \(g(x)=x^2+1\), which one grows faster?

In this case, both functions \(f(x)\) and \(g(x)\) will grow as fast as each other. This is the case for big values of \(x\), as the values of the other terms in the functions (\(3x\) and \(1\)) will not affect significantly the values of \(f(x)\) and \(g(x)\), as much as the value of \(x^2\) does.

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

\(10,000\) | \[\begin{align}f(10,000) &= 10,000^2+3 \cdot 10,000 \\&= 100,030,000\end{align}\] | \[\begin{align}g(10,000) &= 10,000^2+1 \\&= 100,000,001\end{align}\] |

As you may understand by now, there are multiple types of growth. There are also different types of models that you can use to model growth, depending on whether the type of data being modelled is **discrete or continuous**. In this article, we'll be exploring two **discrete models (linear and geometric)**, and two **continuous models (exponential and logarithmic)**.

Discrete data refers to things that you can count, and can only take specific values.

A **model of growth** is considered **discrete** when the values in the model change at specific intervals and not continuously.

Discrete models of growth include linear and geometric models. Let's now explore the characteristics of each one.

**Linear**** growth** is where an initial value or quantity increases or decreases by a **fixed amount** per unit of time, that is, a constant number gets added to the initial value. For example, if a population increases by 100 each year.

The **formulas for linear growth** are as follows:

Type of growth | Type of formula | Formula | Description |

Linear | Explicit | \[ P_n = P_0 + d \cdot n, \] | \(P_n\Rightarrow\) is the quantity after \(n\) time intervals, \(P_0\Rightarrow\) is the initial quantity, \(d\Rightarrow\) is the increase in quantity per time interval, that is, the fixed amount by which the quantity increases in each time interval, \(n\Rightarrow\) is the number of intervals since the initial quantity. |

Recursive | \[ P_n = P_{n-1} + d, \] | \(P_{n-1}\Rightarrow\) is the quantity after \(n-1\) time intervals. |

If you use the **recursive formula of linear growth** to predict future values, for example, after 7 years, then you will have to calculate all the values in the previous years beforehand. In this case, it is more convenient to use the **explicit formula.**

Let's go back to the scenario from the introduction of this article. You have \($1000\) now, with an extra \($100\) added at the end of each year.

**a)** Form an equation to determine the amount of money you will have in the year \(n\).

**b)** Calculate how much money you will have in \(5\) years.

**c) **Using your equation, determine how many years it will take you to reach \($2,500\) .

**Part a****:**

We know that the initial amount of money \(P_0=$1,000\), and the increase in quantity per time interval \(d\) is the **fixed amount** \($100\).

Using the explicit formula of linear growth, we can say that the amount of money that you will have in \(n\) years is given by:

\[ P_n = 1,000 + 100 \cdot n, \]

where \(n\) is the number of years.

**Part b:**

To calculate how much money you will have in \(5\) years, you can use the equation above, and substitute \(n=5\).

\[\begin{align}P_n &= 1,000 + 100 \cdot n \\\newline P_5 &= 1,000 + 100 \cdot 5 \\\newline &= 1,000 + 500 \\\newline &= 1,500\end{align}\]At the end of year 5 you will have \($1,500\).

In the same way, we can calculate the amount of money that you will have after \(1, 2, 3,\) and \(4\) years, and plot the values to see what the graph of the linear growth model would look like. We will use the **recursive formula** in this case, just to show you how it works:

\[\begin{align}P_n &= P_{n-1} + d, \\\newline P_1 &= P_0 + 100 = 1,000 + 100 = 1,100 \\\newline P_2 &= P_1 + 100 = 1,100 + 100 = 1,200 \\\newline P_3 &= P_2 + 100 = 1,200 + 100 = 1,300 \\\newline P_4 &= P_3 + 100 = 1,300 + 100 = 1,400\end{align}\]The **graph of the linear growth model** is the following:

Notice that the points in the graph grow by the same amount each year, therefore, if you draw a line crossing all the points you will get a **straight line**.

**Part c****:**

To find how many years it will take you to reach \($2,500\), we say that \(P_n=2,500\), then solve for \(n\).

\[\begin{align}2,500 &= 1,000 + 100 \cdot n \\\newline 100 \cdot n &= 2,500 - 1,000 \\\newline 100 \cdot n &= 1,500 \\\newline n &= \frac{1,500}{100} \\\newline n &= 15\end{align}\]It will take you 15 years to reach \($2,500\).

**Geometric growth** is when the initial value or quantity increases or decreases as a **ratio** to the existing quantity. For example, if a population grows by \(10\%\) each year.

The **formulas for geometric growth** are described in the table below:

Type of growth | Type of formula | Formula | Description |

Geometric | Explicit | \[ P_n = P_0(1+r)^n, \] | \(P_n\Rightarrow\) is the quantity after \(n\) time intervals, \(P_0\Rightarrow\) is the initial quantity, \(r\Rightarrow\) is the rate at which the quantity is growing ( \((1+r)\Rightarrow\) is the \(n\Rightarrow\) is the number of time intervals. |

Recursive | \[ P_n = (1+r)P_{n-1}, \] | \(P_{n-1}\Rightarrow\) is the quantity after \(n-1\) time intervals |

For geometric growth, the number of time intervals \(n\) is a positive integer.

Notice that the explicit formula of geometric growth is in the form of \(y=a \cdot b^x\). This means the y-intercept in this form of growth (like for linear growth) is the initial population size.

The points in the **graph of a geometric growth model** have a characteristic shape. They are not lined up in a straight line, because the growth is based on a percentage. Therefore, the larger the quantity, the larger the growth will be, starting with a slow growth, and then rising up from left to right. We will see this more clearly in the examples below.

The frog population in a pond grows \(20\%\) each year. There are currently \(25\) frogs in the pond.

**a)** Find an equation for the number of frogs after \(n\) years and graph it.

**b)** Determine the frog population size after \(15\) years.

**Part a**:

The initial population of frogs is \(25\). Therefore, \(P_0=25\).

If the population grows \(20\%\) each year, \(20\%\) of the previous population size gets added in a time interval. Hence, \(r=0.2\).

Using the **explicit formula of geometric growth**, we can say that the number of frogs after \(n\) years is given by:

\[\begin{align}P_n &= 25(1+0.2)^n \\\newline P_n &= 25(1.2)^n,\end{align}\]where \(n\) is the number of years, and the **growth factor** is \(1.2\).

**Part b:**

To calculate the frog population size after \(15\) years, you can use the equation above, and substitute \(n=15\).

\[\begin{align}P_{15} &= 25(1.2)^{15} \\&= 385\end{align}\]

We can say that the frog population after \(15\) years will be \(385\).

To be able to graph the geometric growth model of the frog population in the pond, we need a few more values. Let's calculate the number of frogs from year \(1\) to \(5\), and also at years \(8\) and \(12\).

\[ \begin{align}P_n &= 25(1.2)^n, \\\newline P_1 &= 25(1.2)^1 = 30 \\\newline P_2 &= 25(1.2)^2 = 36 \\\newline P_3 &= 25(1.2)^3 = 43 \\\newline P_4 &= 25(1.2)^4 = 52 \\\newline P_5 &= 25(1.2)^5 = 62 \\\newline P_8 &= 25(1.2)^8 = 107 \\\newline P_{12} &= 25(1.2)^{12} = 223\end{align}\]Now let's see what the graph looks like:

Notice the shape of the graph rising up from left to right as the population gets larger.

**Compound interest** is another type of geometric growth model, where interest is paid on an investment and on any interest earned already.

This idea of compound interest might sound familiar to you, from the second option in the introductory scenario of this article. That option was to put the \($1,000\) in a bank account that pays \(6\%\) annual interest compounded semi-annually. Let's work out the how much you will earn after \(5\) years to decide which option is more profitable for you.

\[\begin{align}A &= 1,000(1+\frac{0.06}{2})^{2 \cdot 5} \\\newline A &= 1,000(1.03)^{10} \\\newline A &= 1,344\end{align}\]After \(5\) years you will have earned \($1,344\), which is less than the \($1,500\) that you would earn with the first option. However, we know that geometric growth rises rapidly as the quantity gets larger. So, let's calculate how much you will have earned after \(15\) years and compare that with the first option \(($2,500)\), to help us make our decision.

\[\begin{align}A &= 1,000(1+\frac{0.06}{2})^{2 \cdot 15} \\\newline A &= 1,000(1.03)^{30} \\\newline A &= 2,427\end{align}\]

According to our results, the first option is indeed better than the second one, at least for the first \(15\) years.

Continuous data refers to things that you can measure and can take infinite values.

A **model of growth** is considered **continuous** when the values in the model change continuously and not at specific intervals.

Continuous models of growth include the exponential and logarithmic functions. We will now describe each one in turn focus on the way they grow.

**Exponential growth models** are basically the same as geometric growth models, where the initial value or quantity increases or decreases as a ratio to the existing quantity, but exponential growth deals with continuous data, not discrete.

Although both types of growth use the same equation, exponential growth makes for a much **smoother graph**, as all the numbers in between the integer values of are accounted for.

The **exponential growth function **is in the form

\[y=a \cdot b^x\]

where:

\(a\Rightarrow\) is the initial value, and \(a \neq 0\),

\(b\Rightarrow\) is the **growth factor**, and \(b > 1\),

The exponential growth function has \(y=0\) as a horizontal asymptote. See the graph below.

The **domain** and **range **of the exponential function can be defined as follows:

Function | Domain | Range |

Exponential | All real numbers | \(y > 0\) |

Graph the function \(f(x)=2^x\).

As you can see, the graph rises up from left to right, rising more rapidly as the quantity gets larger. The growth is determined by the value of \(b\), in this case, \(b=2\).

Read about Exponential Growth and Decay for more in-depth details and examples.

The **logarithmic function** is the inverse of the exponential function.

That is, if \(y = b^x\) then \(log_b y = x\).

If \(2^3 = 8\), then \(log_2 8 = 3\). This is read as the log base \(2\) of \(8\) equals \(3\).

The logarithmic function has \(x=0\) as **vertical asymptote**, and its **domain** and **range** are as follows:

Function | Domain | Range |

Logarithmic | \(x > 0\) | All real numbers |

Now let's look at the shape of the logarithmic function graph in the example below.

Graph the function \(f(x)=log_2 x\).

In this case, the value of the function increases rapidly at first, and then it slows down its growth rate.

Read our explanations about Logarithmic function, and Evaluating and graphing logarithmic functions to expand your knowledge on this topic.

In exponential functions, when a quantity increases at a multiplicative rate each time interval, the quantity after \(t\) time intervals can be obtained with the formula in the form:

\[ y = a(1+r)^t, \]

where:

\(a\Rightarrow\) is the initial quantity,

\(r\Rightarrow\) is the rate at which the quantity is growing (**growth rate**) as a decimal,

\((1+r)\Rightarrow\) is the growth factor, also known as growth multiplier or common ratio,

\(t\Rightarrow\) is the number of time intervals.

Notice that \( y = a(1+r)^t \) is equivalent to the formula \( P_n = P_0(1+r)^n\), just using different variables.

Let's see an example.

A property bought in the year \(2000\) for \($170,000\) has increased in value to \($250,000\) by the year \(2010\).

**a)** Calculate the growth rate between the years \(2000\) and \(2010\).

**b)** What is the growth rate resulting from questions \(a\) expressed as a percentage.

**Part a:**

We know that the initial amount \(a = $170,000\), and the numbers of years between \(2000\) and \(2010\) is \(10\). Therefore, \(t = 10\) and the amount after \(10\) years is \(y = $250,000\).

Now we can substitute those values in the formula \( y = a(1+r)^t\), and solve for \(r\).

\[\begin{align}y &= a (1 + r)^t \\\newline 250,000 &= 170,000 (1 + r)^{10} \\\newline \frac{250,000}{170,000} &= (1 + r)^{10} \\\newline \frac{25}{17} &= (1 + r)^{10} \\\newline (\frac{25}{17})^\frac{1}{10} &= [(1 + r)^{10}]^\frac{1}{10} \qquad \text{ raise both sides by } \frac{1}{10} \\\newline 1.0393 &= 1 + r \\\newline r &= 1.0393 - 1 \\\newline r &= 0.0393\end{align}\]**Part b:**

To express the growth rate \(r = 0.0393\) as a percentage, we simply need to multiply by \(100\).

\[r = 0.0393 \cdot 100 = 3.93\%\]

- Growth is a value of increase or decrease in relation to a previous value or an initial value.
- The growth factor can be a constant or a rate, depending on which type of growth we are dealing with.
- Discrete models of growth include linear and geometric, and continuous models include logarithmic and exponential growth.
- Linear growth increases by a constant number each time interval.
- Geometric and exponential growth increase at a constant rate of a previous quantity.
- Exponential and geometric growth differentiate by using real numbers or integers respectively, resulting in smoother graphs for exponential growth.
The logarithmic function is the inverse of the exponential function.

More about Growth of Functions

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