In the previous guide, we learned how to find the relative or local extrema of a function using derivatives—if the point is a critical point and the derivative of the function is negative to its left and positive to its right, it is a relative minimum (vice versa it is a relative maximum). Naturally one may wonder, can we use derivatives to find the absolute or global extrema of a function? The answer is yes, we can! 😁

🔎 Finding Absolute Extrema

To find the absolute extrema of a function on a closed interval, one should apply the Candidates Test. Remember from key topic 5.2 that absolute extrema are the maximum and minimum values of a function over its entire domain, rather than a specific interval.

Something to keep in mind for absolute extrema is that they can only occur at critical points or at endpoints of a function on a closed interval.

🪜 Candidates Test Steps

The following steps outline the process for using the Candidates Test to determine absolute extrema:

1. 🧐 Find the critical points of the function. Remember, the critical points of a function are the points where the first derivative of the function is equal to zero or undefined.
2. ✏️ Evaluate the function at the critical points. This will give you the y-values of the critical points.
3. 🔚 Evaluate the function at the endpoints of the interval. This will get you the y-values of the endpoints.
4. 🍎🍏 Compare the y-values of the critical points and endpoints to determine the absolute extrema. The largest y-value is the absolute maximum, and the smallest y-value is the absolute minimum.

The Candidates Test is particularly useful when a function cannot be expressed in closed form, or when it is difficult to find the critical points. You can think of yourself as a detective, evaluating the function at specific points, called candidates, and then putting your data together to determine absolute extrema. 🕵🏿

📝 Candidates Test Walkthrough

Let’s walk through an example together. Let’s use the following function:

At what $x$ does the absolute maximum of $h$ over the closed interval $[-4,0]$ occur?

Following the steps of the Candidates Test as shown above:

🧐 Step 1: Find Critical Points

We want to first find the critical points of the function on the interval. Let’s take the derivative of the function and set it equal to 0.

Since $h'(x)$ is defined at every $x$, we simply need to find the points where $h'(x)=0.$

Based on this work and factoring out the GCF, we know we have critical points at $x=-2,0$. Thus, these two points and are two candidates for the absolute maximum of $h(x)$ on the interval $[-4,0]$.

✏️ Step 2: Evaluate Critical Points

Next, we need to evaluate $h(x)$ at $x=-2,0.$

🔚 Step 3: Evaluate Endpoints

The third step is to evaluate the function at the endpoints of the given interval $[-4,0]$.

We have already evaluated $h(0)$ above!

🍎🍏 Step 4: Compare Outputs

The final step is to compare all these y-values with each other to find the largest one, since the question is asking us for the absolute maximum. In this case, $h(-2)=20$ is the largest and is thus the absolute maximum.

Therefore, the absolute maximum of $h$ over the interval $[-4,0]$ occurs at $x=-2$.

📝 Candidates Test Practice Problems

Now that you are a pro, it’s time to do some practice on your own!

❓ Candidates Test Problems

Candidates Test Question 1:

Let $h(x)=-x^4+2x^2$. At what $x$ does the absolute maximum of $h$ over the closed interval $[0,2]$ occur?

Candidates Test Question 2:

Let $h(x)=-x^{5}+5x$. At what $x$ does the absolute minimum of $h$ over the closed interval $[-2,1]$ occur?

✅ Candidates Test Solutions

Candidates Test Question 1:

🧐 Step 1: Find Critical Points

Following the steps of the Candidates Test, we want to first find the critical points of the function on the given interval.

Since $h'(x)$ is defined at every $x$, we simply need to find the points where $h'(x)=0.$

Now we have critical points at $x=-1,0,1$. Do all of these x-values fit in the given closed interval $[0,2]$? Nope! Only $0$ and $1$ fall in the relevant interval and thus are the only critical points that are candidates for the absolute maximum of the function on the interval $[0,2]$.

✏️ Step 2: Evaluate Critical Points

Next, we need to evaluate $h(x)$ at $x=0,1.$

🔚 Step 3: Evaluate Endpoints

The third step is to evaluate the function at the endpoints.

We have already evaluated $h(0)$ above: $h(0)=0$.

🍎🍏 Step 4: Compare Outputs

The final step is to compare all these y-values with each other to find the largest one to find the absolute maximum. In this case, $h(1)=1$ is the largest and is thus the absolute maximum.

Therefore, the absolute maximum of $h$ over the interval $[0,2]$ occurs at $x=1$.

Candidates Test Question 2:

🧐 Step 1: Find Critical Points

Following the steps of the Candidates Test, we want to first find the critical points of the function on the given interval $[-2,1]$.

Since $h'(x)$ is defined at every $x$, we simply need to find the points where $h'(x)=0.$

Therefore, the critical points are at $x=-1,1$. Since both fall in the relevant interval, they are both candidates for the absolute minimum of the function on the interval $[-2,1]$.

✏️ Step 2: Evaluate Critical Points

Next, we need to evaluate $h(x)$ at $x=-1,1.$

🔚 Step 3: Evaluate Endpoints

The third step is to evaluate the function at the endpoints.

We have already evaluated $h(1)$ above: $h(1)=4$.

🍎🍏 Step 4: Compare Outputs

The final step is to compare all these y-values with each other to find the smallest one to find the absolute minimum. In this case, $h(-1)=-4$ is the smallest value and is thus the absolute minimum.

Therefore, the absolute minimum of $h$ over the interval $[-2,1]$ occurs at $x=-1$.

⭐ Closing

In conclusion, the Candidates Test is a useful tool for determining the absolute extrema of a function. It involves finding the critical points, evaluating the function at those points and at the endpoints, and comparing the y-values to determine the absolute extrema.

You got this!