Use the graphs of ƒ' and ƒ" to complete the following ste...

Question

Use the graphs of ƒ' and ƒ" to complete the following steps.

Plot a possible graph of f.

Accepted Answer

Hi, everyone. Let's take a look at this practice problem. This problem says give the graphs of F and F, which among these graphs is a possible graph of F. Below the question, we're given a graph that shows our 1st and 2nd derivatives of F of X. So we have F of X and F of X. For F of X, we see that it crosses the x-axis at X equal to -12, X equal to 0, and X equals to 12. And our function is concave up, when X is less than 0 and it's concave down when X is greater than 0. For our second derivative F of X, we see that it crosses the x-axis at X equal to -5 and X equal to 5, and this function is concave down everywhere. We're also given 4 possible choices as our answers, and each one of these answer choices. Contains a graph of a possible F of X function. And we'll come back to those graphs at the end. So, in order to determine which of our following graphs, Is the correct graph, we need to determine some properties of F of X based upon our graphs for FX and F of X. So starting to look at FX, we see that F of X is equal to 0. At X equal to -12, X equal to 0, and X equal to 12. So these are going to be our critical points, since our critical points are located whenever our first derivative is equal to 0, as well as when our first derivative is undefined. However, we have a very smooth curve, so it looks like it's defined everywhere along the x-axis. So these are going to be our critical points, and we need to check to see if we have a local minimum or maximum at these um critical points. So we're gonna start off by looking at X equal to minus 12, and When X is equal to -12, we can check to see if we have a local minimum or maximum by using the second derivative test. So if our second derivative is negative, that means that we have a local maximum, and if our second derivative is positive, that means we have a local minimum. So at X equal to -12, our second derivative FX actually has a negative value. This means that we have a local maximum. At X equal to minus 12. If we look at X equal to 12, we have the same thing. Again, our second derivative, FX is negative when X is equal to 12. So therefore, we have a local maximum. We have local maximums at the points of X equal to -12 and X equal to 12. However, if we look at X equal to 0, we'll see that we have a local minimum. And that's because at X equal to 0, our second derivative FX actually has a positive value. And since we have a positive value, that means we have a local minimum. We could have also used the first derivative test to determine whether these points were minimum, local minima or maxima, and we'd get the same result. So now we'll move on to our second derivative, and we need to determine if we have any inflection points. So those occur when our second derivative F of X is equal to 0, and this occurs at X equal to minus 5 and X equal to 5. So, in order to have an inflection point, we need to have the concavity of our function F to be changing. And here, if we look at X equal to -5, we see that our second derivative is going from a negative value to a positive value. That means that our function is going from concave down to concave up. So here we do have a change in our concavity, so we do have an inflection point at X equal to -5. Likewise, if we look at X equal to 5, we see that our second derivative is changing from a positive value to a negative value. So it's changing from concave up to concave down. So again, we have a change in our concavity, so we do have an inflection point at X equal to 5. Therefore, we have inflection points. At X equal to. -5. And X equal to 5. So now we need to use this information to determine which of the graphs in our choices is the correct choice. So, we'll start off by looking at choice A, and here in our graph, we see that we have a local minimum at X equal to -12. However, we are looking for a local maximum at X equal to -12, so that means choice A cannot be correct. If you look at choice B, we have a local maximum at X equal to -12, as well as X equals to 12. So that corresponds to what we found earlier based upon our first derivative. We also have a local minimum at X equal to 0, again, corresponding to what we found based upon the first derivative of our original graph. We also see that we have inflection points at X equal to -5 and X equal to 5. So at X equal to -5, we're changing from concave down to concave up, and at X equal to 5, we're changing from concave up to concave down, and that matches what we found earlier based upon our second derivative. So, since all of these properties match, we see that the correct answer choice actually corresponds to answer choice B. Now, we can look at the last two answer choices, Ansho C and D, to see why they are not correct. If you look at answer choice C, we see that we have local maxima at X equal to -12 and X equal to 12, but we do not have a local minimum at X equal to 0, so that's why C is incorrect. And for choice D, again, if we look at X equal to -12, we have what appears to be a local minimum. However, we are looking for a local maximum. So that's the reason why choice D is not correct. So the correct answer choice for this problem is actually answer choice B. So, I hope this explanation has been useful, and I'll see you in the next video.

Use the graphs of ƒ' and ƒ" to complete the following steps. <IMAGE>
Plot a possible graph of f.

Key Concepts

Derivative and Its Graph

Second Derivative and Its Graph

Critical Points and Inflection Points

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