Identifying Extrema

In Exerc...

Question

Identifying Extrema

In Exercises 19–40:

a. Find the open intervals on which the function is increasing and those on which it is decreasing.

b. Identify the function’s local extreme values, if any, saying where they occur.

g(x) = x√8 − x²

Accepted Answer

Hello. In this video, we are given the function F of X is equal to X, multiplied by the square root of 16 minus X. We are going to describe the intervals of which F of X is increasing and decreasing, and identify the local extreme. Now, in order to find the intervals of increasing and decreasing, and to find the local extrema, the first thing we need to do is we need to take the derivative of the function. Before we take the derivative, let's go ahead and represent the function in an exponential form. We can rewrite the given function as F of X is equal to X multiplied by the quantity, 16 minus X raised to the half power. Now what we are going to do is we are going to take the derivative by using the product rule. That will give us X multiplied by 1/2, multiplied by the quantity 16 minus X, raise the negative half power, multiplied by -1, plus 16 minus X rates the half power multiplied by 1. This will further simplify to be negative X divided by 2 multiplied by the square root of 16 minus X, plus the square root of 16 minus X. And by combining the fractions together, we can rewrite the derivative to be F of X is equal to negative X plus 2 multiplied by 16 minus X, all divided by 2 multiplied by the square root of 16 minus X. Now, what we want to do is you want to go ahead and find the critical points to this derivative. In order to find the critical points, we are going to take the value or find the values of X that cause the numerator to be zero. In order to do this, we will take the numerator, set it equal to 0, and solve for X. That is going to give us negative X plus 2 multiplied by the quantities 16 minus X equal to 0. By distributing 2 into the quantity, we get negative X plus 32 minus 2 X is equal to 0. And this will further simplify to be -3 X plus 32 equal to 0. Finally, by sobbing for X, we get X's equal. To 32 divided by 3, and this has an approximate value of 10.67. Now what we want to do is we want to determine the intervals of increasing and decreasing. 01 more thing we need to do as well, we need to determine the domain of the original function. Now for the domain Since we have a square root function, we want to make sure that the square root is always positive and not negative. So that means that we are going to take 16 minus X, set it greater than it equal to 0, and solve. And this tells us that 16 must be greater than or equal to X. What this means is that X can be any number, but the largest number it can be is 16, so the domain will be from negative infinity to 16. OK, now that we've discussed the domain, and we have the critical 0.32 divided by 3, let's go ahead and use the first derivative test to define or determine the intervals of increasing and decreasing. So, we have 32 divided by 3, with the highest value being 16 and the lowest value being negative infinity. Let's go ahead and take some testing points to plug into the derivative. Some testing points we can take is negative or 9. And 11. F9 gives us the value of 0.94, which is positive. So the graph will be increasing from negative infinity to 32 divided by 3. And the value of F of 11 will give us the output of negative 0.22. Since this is negative, the graph will be decreasing from 32 divided by 3 to 16. So, we just determined that we have the intervals of increase. From negative infinity. To 32 divided by 3. And the graph will be decreasing. From 32 divided by 3. 216. This is going to be the first part of the solution. Now, the second part of the solution wants us to find the local extreme values. Well, since we know that as we pass through the critical point, the graph will be shifting from increasing to decreasing, which represents the behavior of a local maximum. So this means that we have a local max. At the value, X is equal to 32 divided by 3. And in order to find the local extreme value, we need to plug in this value, X is equal to 32 divided by 3 into the original function. And that is going to give us F of 32 divided by 3 is equal to 128 multiplied by the square root of 3 divided by 9, and this is going to be the final solution to the problem. So I hope this video helps you in understanding how to approach this problem, and I will go ahead and see you all in the next video.

Identifying Extrema

In Exercises 19–40:

a. Find the open intervals on which the function is increasing and those on which it is decreasing.

b. Identify the function’s local extreme values, if any, saying where they occur.

g(x) = x√8 − x²

Key Concepts

Derivative and Critical Points

Increasing and Decreasing Intervals

Local Extrema

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