In Exercises 21-36, find the absolute maximum and minimum...

In Exercises $21-36,$ find the absolute maximum and minimum values
of each function on the given interval. Then graph the function. Identify the points
on the graph where the absolute extrema occur, and
include their coordinates.
$$
f(x)=\frac{2}{3} x-5, \quad-2 \leq x \leq 3
$$

Key Concepts

Extreme Value Theorem

This theorem states that every continuous function on a closed interval will have both an absolute maximum and an absolute minimum value somewhere in that interval. It provides the foundational guarantee for finding extrema in calculus.

Evaluating Endpoints

When identifying the absolute extrema of a function on a closed interval, it is crucial to evaluate the function at the interval's endpoints. This is because even if there are no critical points within the interval, the extreme values may occur at these endpoint locations.

Monotonic Functions

Monotonic functions consistently increase or decrease over their domain, which makes it easier to determine where the absolute extreme values will occur. For instance, in an increasing function, the absolute minimum is found at the left endpoint, while the absolute maximum is at the right endpoint.

Graphing Linear Functions

Graphing a linear function involves plotting a straight line determined by its slope and y-intercept. Understanding how to represent the function graphically is linked to identifying key points such as the extreme values and their corresponding coordinates on the graph.

Transcript

00:01 Consider the following function f of x equal to 2 over 3x minus 5, for x ranging between minus 2 and 3.

00:10 We want to find the absolute maximum and minimum values of the function, then graph the function and identify on the graph where these extreme points occur.

00:22 So to find the extreme points, the first thing we want to do is find the critical points.

00:26 We find the critical points by evaluating the first derivative.

00:28 So let's calculate f prime of x.

00:30 We obtain 2 thirds, corresponding to the derivative of 2 over 3x, because the derivative of minus 5, the constant, is equal to zero and has zero contribution to our slope.

00:50 So now to find the critical points, we want to find the values of x for which f prime of x is equal to zero.

01:03 And for our given problem, for f prime of x equal to the constant 2 thirds, we see that this is never the case, meaning that we don't have an extreme point, or we don't have a critical point on our graph for x ranging from minus infinity to plus infinity.

01:19 But because of our closed interval, we may have an extreme point at our endpoints.

01:24 Now let's see what happens at our endpoints.

01:31 Meaning by, i want to check the values of f when it is evaluated at our lower boundary and our upper boundary.

01:40 And just by looking at f prime of x, i know that this is a constant slope.

01:47 We are dealing with a linear function that is always increasing, which means that intuitively, i'm expecting the minimum to occur at our lower value of x and the maximum to occur at the maximum value of x, just because our curve is always increasing.

02:06 But let's verify this.

02:07 Oops, excuse me.

02:08 Let's calculate f of minus 2, and we obtain minus 4 over 3, minus 5, yielding minus 19 over 3.

02:29 And f of 3 will give us 2 minus 5, yielding minus 3...

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