Identify the inflection points and local maxima and minima of ...

Question

Identify the inflection points and local maxima and minima of the functions graphed in Exercises 1–8. Identify the open intervals on which the functions are differentiable and the graphs are concave up and concave down.
7. y=sin|x|, -2π≤x≤2π

7. y=sin|x|, -2π≤x≤2π

Accepted Answer

Hello there. Today we're gonna solve the following practice problem together. So, first off, let us read the problem and highlight all the key pieces of information that we need to use in order to solve this problem. Determine the inflection points and intervals. Of concavity for the function H of X is equal to the absolute value of X to the power of 3, where 2 is less than or equal to X and X is less than or equal to 2. Awesome. So it appears for this particular problem, we're asked to solve for two separate answers. We're asked to determine the inflection points and the intervals of concavity for the specific function of H of X, given this particular domain, which once again is negative 2 is less than or equal to X and X is less than or equal to 2. OK. So, with that said, now that we know we're ultimately trying to solve for. Once again, we're asked to determine the inflection points and intervals of concavity for the function H of X is equal to, so let's write that down for our function H of X is equal to the absolute value of X to the power of 3 on our interval, which I said domain, but specifically on the interval, -2, which is less than or equal to X and X is less than or equal to 2. Awesome. So we need to follow these steps to help us determine our final answers. So our first step is we need to consider the definition of the function. So as we should recall a note, the function H of X is equal to the absolute value of X to the power of 3. is a piecewise function, and we must break it into two cases based on whether X is positive or a negative value. So with that said, let's make a note. So when X is greater than or equal to 0, that will mean that the absolute value of X is equal to X. So that will mean that H of X is equal to X to the power of 3. However, if X is less than 0, that will mean that the absolute value of X is going to be equal to negative X. So that will mean that H of X is going to be equal to parentheses X to the power of 3, which will be equal to negative X to the power of 3. Thus, therefore, as a result, we can rewrite our function as H of X is equal to curly bracket of X to the power of 3. If X is greater than or equal to 0 and negative X to the 3, if X is less than 0, fantastic. So with that said, our next step now, step 2, is we need to determine or figure out our first derivative, and after that, next we are next is we're going to have to calculate the first derivative for. Case. So, with that said, we need to focus our attention on X is greater than or equal to 0. So that will mean that H of X is equal to H of X is equal to X to the power of 3. So, that will mean that H X is equal to 3 x 2. Once again, this is 4, X is greater than or equal to 0. So for the condition where X is less than 0, that will mean that H of X is equal to negative X to the power of 3. So that will mean that H of X, where once again this prime symbol denotes the first derivative. So H of X will be equal to -3X is the power 2. Once again, this is for X is less than 0. So X is less than 0. Once again, moving right along now, our next step now, which is our 3rd step, we need to find the 2nd derivative now. So, let us now compute the 2nd derivative for each condition. So, for X is greater than or equal to 0, you know that H of X is equal to 3 x to the power of 2. So that will mean that H of X, which once again this time is 2 prime symbols, so it's the second derivative. So H of X is equal to 6 X, and once again, this is for the case, X is Gonna be greater than 0. So this is for the case once again where X is greater than 0. Once again, greater than or equal to 0. OK. So now, our second condition. when X is less than 0, as we should recall, H of X is equal to -3 x 2. So that will mean that H of X will be equal to -6 X. And once again, this is for the condition where X is less than 0. Awesome. Moving right along here. So, our 4th step now. We need to determine the intervals of concavity, so we need to analyze the sign of H of X, or second derivative of H of X, to determine the concavity. So, with that said, for when X is Greater than 0, let us note that H of X is equal to 6 X, and because it's positive, that will mean that the function is concave up. For the condition where X is less than 0, H is a positive value because X is negative, so 6 X. is positive, so therefore, the function will also be concave up, which I'm denoting as CU for short. Awesome. So once again, for the condition, X is less than 0, H is positive because X is negative, so -6X is positive, so the function is concave up as well. So therefore, without a doubt, the function will be. Concave up on the closed interval, -2.2 in square brackets because square brackets indicate a closed interval. Awesome. So with that said, we've officially solved for all of our final answers. So let's go look at the problem itself to quickly recap everything that we've learned. So that means without. So without a doubt, our final two answers are for the inflection points there are none, because the inflection points we have none, because while H 0 is equal to 0, the concavity does not change from positive to negative or vice versa. So that means that there are no inflection points, and our answer for the concavity will be, it's going to be concave up on square brackets -22. And that's it, we've officially solved for this problem. Thank you so much for watching. Hopefully that helped, and I can't wait to see you in the next video. Bye.

Identify the inflection points and local maxima and minima of the functions graphed in Exercises 1–8. Identify the open intervals on which the functions are differentiable and the graphs are concave up and concave down.
7. y=sin|x|, -2π≤x≤2π

Key Concepts

Inflection Points

Local Maxima and Minima

Concavity and Differentiability

Watch next

Identify the inflection points and local maxima and minima of the functions graphed in Exercises 1–8. Identify the open intervals on which the functions are differentiable and the graphs are concave up and concave down.7. y=sin|x|, -2π≤x≤2π

Key Concepts

Inflection Points

Local Maxima and Minima

Concavity and Differentiability

Watch next

Identify the inflection points and local maxima and minima of the functions graphed in Exercises 1–8. Identify the open intervals on which the functions are differentiable and the graphs are concave up and concave down.
7. y=sin|x|, -2π≤x≤2π