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! 😁

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🔎 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:

$$h(x)=x^3+3x^2+16$$

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.

$$h'(x)=3x^2+6x$$

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

$$0=3x^2+6x$$ $$0=3x(x+2)$$

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 $x=-2,0.$

$$h(-2)=20$$ $$h(0)=16$$

🔚 Step 3: Evaluate Endpoints

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

$$h(-4)=0$$ $$h(0)=16$$

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 $x=-2$.

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📝 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.

$$h'(x)=-4x^3+4x$$

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

$$0=-4x^3+4x$$ $$0=-4x(x^2-1)$$ $$0=-4x(x+1)(x-1)$$

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 $x=0,1.$

$$h(0)=0$$ $$h(1)= 1$$

🔚 Step 3: Evaluate Endpoints

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

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

$$h(2)=-8$$

🍎🍏 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 $[-2,1]$.

$$h'(x)=-5x^4+5$$

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

$$0=-5x^4+5$$ $$0=-5(x^4-1)$$ $$0=-5(x^2+1)(x+1)(x-1)$$

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 $x=-1,1.$

$$h(-1)=-4$$ $$h(1)= 4$$

🔚 Step 3: Evaluate Endpoints

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

$$h(-2)=22$$

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 $x=-1$.

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⭐ 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!