Absolute Extrema on Finite Closed Intervals

Question

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.

h(x) = ³√x, −1 ≤ x ≤ 8

Accepted Answer

Hi, everyone, let's take a look at this practice problem. This problem says determine the absolute minimum and maximum values of the function F of X, which is equal to 3, multiplied by the cube root of X on the closed interval from 8 to 1 and identify where they occur. So this problem we need to find the absolute minimum and maximum values of our function F of X. So the first thing we want to do is identify the critical points, and to do that, we'll need to take a derivative of our function F of X. So we'll calculate F of X, and that's going to be equal to the derivative with respect to X. Of the quantity of 3 multiplied by x raised to the 1/3 power in quantity, and here we've rewritten the q root of x as x raised to the 1/3 power, and this will make taking the derivative a bit easier. So to take this derivative, we'll just need to use our constant multiple rule. And so taking this derivative, this is going to be equal to x raised to the minus 2/3 power, which we can rewrite as being equal to 1 divided by the cube root of X2. So now we need to identify the critical points, and recall that critical points are the X values which make our first derivative either equal to zero or undefined. Now if we look at our function here for F of X, we see that we have a fraction, and in the numerator we actually have a constant of 1. So since we have a constant in the numerator, that means that our first derivative here is never going to be equal to 0. However, we can have our first derivative be undefined whenever the denominator is equal to 0. So we're going to set the cube root of X2 equal to 0. And if we cube both sides, and then take the square root root of both sides, we'll get X is equal to 0 as our critical point. So now we need to determine whether this critical point is a local minimum or maximum, and for this we're going to use the first derivative test. Now, we call, in order for us to have a local extrema at a critical point, we need the sign of our first derivative to actually change. So either going from negative to positive or from positive to negative. So since our numerator is a constant and it's positive, the sign of our first derivatives is going to be totally determined by the sign of the denominator. So, if we look at the cube root of X2d, We see that this quantity is always going to be greater than or equal to 0. Since when we square the X value, that will make it positive, and we take the cube root of a positive value, we get a positive number. And we've already said that we can't have X equal to 0, since our first derivative would be undefined at that point. So, if we look at the points where the cube root of X2 is greater than 0, that's going to be for all values of X, except for X equal to 0. So that means that our function F of X is always going to be greater than 0 whenever it is defined. And since it's always greater than 0, that means it's always positive. So we will have no sign change at X equal to 0, and so therefore, X equal to 0 is not going to be a point of Xtrema. So, in order to find the absolute minimum and maximum values, we now just need to evaluate the end points of our interval. So, we'll start off by looking at X equal to -8. So we'll have F of -8. And that's going to be equal to 3, multiplied by the cube root of -8, and when we evaluate this expression, this is going to be equal to -6. And the other end point that we need to look at is when X is equal to 1, so we'll calculate F of 1, and that's going to be equal to 3, multiplied by the cube root of 1, which is going to be equal to 3. And so these two points are going, these two values are actually going to be our absolute minimum and absolute maximum. So we'll have an absolute Minimum That has a value. Of -6, and this occurs at X equal to -8. And if we look at the absolute. Maximum Its value is 3 at x equal to 1. And so those are the answers that this problem was looking for that we have that the absolute minimum is -6 at X equal to -8, and the absolute maximum is 3 at x equal to 1. So I hope that this explanation has been useful, and I'll see you in the next video.

Key Concepts

Absolute Extrema

Critical Points

Graphing Functions

Watch next