Second Derivative Test Locate the critical points of the follo...

Question

Second Derivative Test Locate the critical points of the following functions. Then use the Second Derivative Test to determine (if possible) whether they correspond to local maxima or local minima.

f(x) = 2x³ - 3x² + 12

Accepted Answer

Hi everyone, let's take a look at this practice problem. This problem says to locate the critical points of F of X, which is equal to 4 XQ minus 12 X2 + 9, and use the second derivative test to identify whether these points are local maxima, minima, or neither. Now, the first part of this question wants us to find the critical points, and so to do that, we first need to take a derivative of our function. We wouldn't need to calculate F of X. And that is going to be equal to the derivative with respect to X. Of the quantity of 4 x cubed. Minus 12 X squad. Plus 9 in quantity. So taking this derivative, this is going to be equal to 12 X squared. 24 X. So recall that for a critical point. Those are the points at which our first derivative is either not defined or where the first derivative is equal to zero. Now, in our case here, our first derivative is a polynomial, and polynomials are defined and continuous everywhere on the real line. That means our first derivative of X is defined everywhere. So in order to find the critical points, we need to just set our first derivative here equal to 0. So we will have 12 X squared. Minus 24 X equal to 0. And we need to just solve this for X. So we can start off by simplifying this expression by dividing both sides by 12. In doing so, I will get X squared, minus 2 X is equal to 0. I can factor out X out of both terms on the left hand side, and I will have X multiplied by the quantity of X minus 2 in quantity is equal to 0. Now we just need to set each of these factors equal to 0. In doing so, we will have X equal to 0 as one critical point, and then we'll have X minus 2 equal to 0, and solving for X, we will get X equal to 2 as our second critical point. So these are our two critical points, X equal to 0 and X equal to 2. So now we need to use the second derivative test to identify whether these points are local maxima, minima, or neither. So we'll need to first calculate our second derivatives, so we'll be looking for F of X. And this is equal to the derivative with respect to X of the quantity, and here we'll use our first derivative, which is going to be 12 x quad minus 24 X. So taking this derivative, this is going to be equal to 24 X minus 24. So we call for the second derivative test, we need to substitute in the values for our critical points and see whether our second derivative is positive or negative at the uh at those points. So we're gonna start off with looking at X equal to 0. We'll calculate F of 0. And this is going to be equal to 24 multiplied by 0 minus 24. And evaluating this expression, this is going to be equal to minus 24. So here we see that our second derivative at X equal to 0 is equal to -24. So it is a negative value. It's so recall for the second derivative test that if the second derivative is negative at a critical point, that means you have a local maximum. So in our case, we have a local maximum at X equal to 0. So now we need the same thing, but 4 X equal to 2. So we'll calculate F of 2. And this is going to be equal to 24, multiplied by 2 minus 24, evaluate this expression, this is going to be equal to 24. So here we see that our second derivative is positive at X equal to 2, which is where our critical point is located. So, by this second derivative test, that means that we have a local minimum. Since our second derivative positive, that gives us a local minimum. So, our final answer for this question is, we're going to have a local maximum at X equal to 0, and a local minimum at X equal to 2. So I hope that this explanation has been useful, and I'll see you in the next video.

Second Derivative Test Locate the critical points of the following functions. Then use the Second Derivative Test to determine (if possible) whether they correspond to local maxima or local minima.

f(x) = 2x³ - 3x² + 12

Key Concepts

Critical Points