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Candidate Test

- Calculus
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Making a movie is, for sure, a big deal. One of the key aspects the writers have to keep in mind is that they need actors for it. But there are way too many actors out there! Doing an audition for everyone would take a long time for sure! What to do then? Well, it is always possible to **narrow the search **under some criteria. This way time and money are saved.

The very idea of narrowing a search can also be used for finding global extrema. Who gives the criteria for narrowing the search? Calculus does, so let's jump into the **Candidates Test** and see how this is done!

A common task found in Calculus is finding the extrema of a function. You can use the First Derivative Test to find the function's critical points and then use the Second Derivative Test to tell whether these critical points correspond to relative maxima or relative minima (or maybe the test is inconclusive). But what about the **absolute ****maximum **and the **absolute ****minimum**? How can you use Calculus to find them?

If you are given a small set of values, let's say 4 numbers, it is an easy task to sort them. This way you can tell which one is the least and which one is the greatest. What about using this same idea for finding absolute extrema? You will first need the set of values, that is, you need **the candidates** to be sorted.

Since the absolute extrema of a function is a key piece to the Candidates Test, it is important to recall what is an absolute maximum and an absolute minimum.

The **a****bsolute maximum,** also known as **g****lobal maximum**, of a function is the greatest output in the range of the function. If \( y_{\text{max}}\) is the absolute maximum of a function \( f(x),\) then \( y_{\text{max}} \geq f(x) \) for all \( x \) in the domain of the function.

The absolute minimum is defined similarly.

The **a****bsolute minimum,** also known as **g****lobal minimum**, of a function is the smallest output in the range of the function. If \( y_{\text{min}}\) is the absolute minimum of a function \( f(x),\) then \( y_{\text{min}} \leq f(x) \) for all \( x \) in the domain of the function.

The absolute maximum and the absolute minimum of a function are known collectively as **absolute extrema.**

Finding the absolute extrema of a function would require you to sort a function's every possible output. This is practically impossible because the function will most likely have an infinite amount of outputs. This is where the Candidates Test comes in handy.

Since one method is included within the other, you might mix up the tape when talking about the Candidates Test and the First Derivative Test.

The Candidates Test is a method for finding the **global extrema** of a function, while the First Derivative Test is a method for finding the **critical points** of a function.

Despite the fact that the critical points are related to **relative** extrema, the First Derivative Test **is used within** the Candidates Test, as you are required to find critical points in order to do the Candidates Test. Let's see why.

When a function is continuous, differentiable, and is defined over a closed interval, there are only two possibilities for each of its absolute extrema:

The absolute maximum/minimum is also a relative maximum/minimum of the function.

The absolute maximum/minimum is at one of the extreme values of the function.

By knowing this, you can narrow the search. This leads to what is known as the Candidates Test.

The Candidates Test is a **method** for finding the absolute maximum and the absolute minimum of a **continuous and differentiable function** that is defined over a **closed interval**. The Candidates Test consists of the following:

- Find all the critical points of the function. This is usually done with the First Derivative Test.
- Evaluate the function at all of its critical points.
- Evaluate the function at both ends of its domain.
- Sort all of the above values. The least value is the global minimum and the greatest value is the global maximum.

Rather than finding every possible output you just focus on the function evaluated at its critical points and at both ends of the function's domain. These values are your **candidates. **Finally, by sorting the candidates, you find the global extrema.

Please note that in order to perform the Candidates Test, you need to be able to find the critical points of the function. If this does not sound familiar at all, or if you need a refresher on the subject, check out our article about the First Derivative Test.

Consider the function

\[f(x) = x^2 -2x +4 \quad \text{ for }\quad 0\leq x \leq 6.\]

Use the Candidates Test to find its absolute extrema.

Answer:

1. *Find all the critical points of the function.*

Begin by performing the First Derivative Test to find the critical points of \( f(x). \)

*Find the derivative of the function.*You can find the derivative of the given function by using the Power Rule, that is\[f'(x)=2x-2.\]*Evaluate the derivative at a critical point \(c\) and set it equal to 0.*This step is rather straightforward. First, evaluate the derivative at \(x=c,\)\[f'(c)=2c-2,\]and then set it equal to 0,\[2c-2=0.\]*Solve the obtained equation for*\(c.\)You can solve the obtained equation by isolating \( c,\) that is\[ \begin{align} 2c-2 &= 0 \\ c &= 1. \end{align}\]

Since the function is defined in the interval \( [0,6],\) and your solution lies within this interval, this is a critical point.

2. *E**valuate the function at all of its critical points.*

You just found that the given function has a critical point at \(x=1,\) now you need to evaluate the function at that point, that is

\[\begin{align} f(1) &= (1)^2-2(1)+4 \\ &= 1-2+4 \\ &= 3. \end{align}\]

3. *Evaluate the function at both ends of its domain.*

The domain of the given function is \( [0,6], \) so you need to evaluate it at \(x=0\) and at \( x= 6.\) That is

\[\begin{align} f(0) &= (0)^2-2(0)+4 \\ &= 0-0+4 \\ &= 4, \end{align}\]

and

\[\begin{align} f(6) &= (6)^2-2(6)+4 \\ &= 36-12+4 \\ &= 28. \end{align}\]

4. *Sort all of the above values.*

\( x \) | \( f(x) \) |

1 (critical point) | 3 |

0 (left extreme value) | 4 |

6 (right extreme value) | 28 |

From the above table, you can conclude that the function has an absolute minimum of 3 located at \( x=1, \)and an absolute maximum of 28 located at \( x=6.\)

To perform The Candidates Test, you need the function to be continuous, differentiable, and to be defined over a closed interval, **otherwise, the test is not guaranteed to work. **

Dive in to see why!

As previously stated, you need a function to be **continuous** and to be defined over a **closed interval **to guarantee that Candidates Test works. Take a look at a discontinuous function.

The function is given as a piecewise function by the following rule:

\[f(x)= \begin{cases} 2-\frac{1}{2} x \quad &\text{ for } \quad 0 \leq x < 2 \\ 5-\frac{1}{2}x \quad &\text{ for } \quad 2 \leq x \leq 4.\end{cases}\]

If you were to perform The Candidates Test you would first find that there are no critical points, so your candidates would be the extreme values of the graph,

\[ f(0) = 2, \]

and

\[ f(4) = 3.\]

This would make you think that the absolute maximum of the function is 3, located at \(x=4.\) However, by looking at the graph you can see that this is not true because \( f(2)=4,\) which is greater than your previous candidate!

And what happens if the function is defined over an **open** interval?

The function rule is now given over an open interval by the following rule:

\[f(x) = x+1 \quad \text{ for } \quad 0 < x \leq 3.\]

Once again, the function has no critical points, so your candidates are the extreme values of the function. The absolute maximum of the function is 4 located at \( x=3.\) However, since the left end of the function is **open **this function **does not** have a global minimum. You can get as close as you want to the left end, but you can never reach it!

Unfortunately, since the Candidates Test is a **method**, there is **no formula** for it. Follow the steps introduced in this article to be proficient in finding global maxima and minima!

Practice use the Candidates Test to find absolute extrema!

Consider the function

\[f(x) = x^3 -3x^2 +1 \quad \text{ for }\quad -2\leq x \leq 4.\]

Use the Candidates Test to find its absolute extrema.

Answer:

1. *Find all the critical points of the function.* You can find the critical points of the function by doing the First Derivative Test.

*Find the derivative of the function.*The given function is a polynomial function, so use the Power Rule to find its derivative, that is \[f'(x)=3x^2-6x.\]*Evaluate the derivative at a critical point \(c\) and set it equal to 0.*Next, write an equation by first evaluating the derivative at a critical point,\[f'(c)=3c^2-6c,\]and then setting it equal to 0, so\[3c^2-6c=0.\]*Solve the obtained equation for*\(c.\)You can solve the above equation by factoring, that is\[ \begin{align} 3c^2-6c &= 0 \\ 3c(c-2) &= 0,\end{align}\]which means that either \(3c=0,\) or \( c-2=0.\) From here you can isolate \(c\) in both cases to obtain the solutions, that is \[c=0,\]and\[c=2.\]

2. *E**valuate the function at all of its critical points.*You found in the previous step that the critical points are \(x=0\) and \(x=2.\) You now need to evaluate the function using these values, so\[\begin{align} f(0) &= (0)^3-3(0)^2+1 \\ &= 1, \end{align}\]

and

\[\begin{align} f(2) &= (2)^3-3(2)^2+1 \\ &= -3. \end{align}\]

3. *Evaluate the function at both ends of its domain.* The domain of the function is \( [-2,4], \) so you need to evaluate the function at \(x=-2\) and \(x=4,\) that is

\[\begin{align} f(-2) &= (-2)^3-3(-2)^2+1 \\ &= -19, \end{align}\]

and

\[\begin{align} f(4) &= (4)^3-3(4)^2+1 \\ &= 17. \end{align}\]

4. *Sort all of the above values.*The values you just found are organized in the following table.

\( x \) | \( f(x) \) |

-2 (left extreme value) | -19 |

0 (critical point) | 1 |

2 (critical point) | -3 |

4 (right extreme value) | 17 |

Sometimes a function will not have critical points. In this scenario, The Candidates Test becomes easier because you only have to evaluate the endpoints of the domain.

Consider the function

\[g(x) = \sqrt{x-1} +1 \quad \text{ for }\quad 1\leq x \leq 5.\]

Use the Candidates Test to find its absolute extrema.

Answer:

1. *Find all the critical points of the function.* As usual, you can find the critical points by using the First Derivative Test.

*Find the derivative of the function.*You can find the derivative of the given function by using the Chain Rule and the Power Rule. Begin by letting \( u(x)= x-1,\) so\[g(u) = \sqrt{u} + 1\]and\[g'(x)=g'(u)u'(x).\]Next, use the Power Rule to find each derivative, that is\[\begin{align} g'(u) &= \frac{1}{2}u^{- ^1/_2}\\ &= \frac{1}{2\sqrt{u}}, \end{align} \]and\[u'(x)=1.\]Finally, substitute back the derivatives into the Chain Rule formula, that is\[ \begin{align} g'(x) &= \frac{1}{2\sqrt{u}} \\ &= \frac{1}{2\sqrt{x-1}}. \end{align}\]*Evaluate the derivative at a critical point \(c\) and set it equal to 0.*First evaluate the derivative,\[ g'(c) = \frac{1}{2\sqrt{c-1}},\]and set it equal to 0, that is\[ \frac{1}{2\sqrt{c-1}}=0.\]*Solve the obtained equation for*\(c.\)There is no \( c\) value that makes the above equation true, so it has no solutions. This means that**the function has no critical point**s.

2. *E**valuate the function at all of its critical points.*You found that the given function has no critical points, so you can skip this step.

3. *Evaluate the function at both ends of its domain.* The domain of the function is \([1,5],\) so evaluate the function at its extreme values, that is

\[ \begin{align} f(1) &=\sqrt{1-1}+1 \\&= 1, \end{align} \]

and

\[ \begin{align} f(5) &=\sqrt{5-1}+1 \\&= 3. \end{align} \]

4. *Sort all of the above values.*

Since you only have two values the sorting is easier. You can conclude that the function has a global minimum of \(1\) located at \( x=1,\) and a global maximum of 3 located at \( x=5.\)

- The Candidates Test is a method used to find the absolute maximum and the absolute minimum of a continuous and differentiable function that is defined over a closed interval.
- The Candidates Test consists of the following:
- Find all the critical points of the function.
- Evaluate the function at its critical points.
- Evaluate the function at both of its extreme values.
- From the above values, the least is the absolute minimum and the greatest is the global maximum.

- It is possible for a function to not have critical points in its domain. In this case, the function is only evaluated at its extreme values.
- Remember that the function needs to be continuous, differentiable, and it needs to be defined over a closed interval. If either of these conditions is not met the test is not guaranteed to work.

The Candidates Test consists of the following:

- Find all the relative extrema of the function.
- Evaluate the function at the extreme values of its domain.
- Sort the above values and the relative extrema of the function. The least value is the global minimum and the greatest value is the global maximum.

**not** the same. However, The First Derivative Test can be used to find the relative extrema of a function, so it might be used while performing The Candidates Test.

There is not a rule for doing the candidate test. There is, however, a series of steps that you can use to do a candidates test.

- Find all the relative extrema of the function.
- Evaluate the function at the extreme values of its domain.
- Sort the above values and the relative extrema of the function. The least value is the global minimum and the greatest value is the global maximum.

More about Candidate Test

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