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.

p(t) = 2t³ + 3t² - 36t

Accepted Answer

Welcome back, everyone. Locate the critical points of F of X equals 4 X cubed plus 18X2 minus 48X and use the second derivative test to identify whether these points are local maxima or minima. So first of all, let's identify the critical points by differentiating the function. Let's identify F of X. We're going to identify the derivative of 4 X cubed plus 18X2 minus 48 X, and we have to apply the power rule. So we're going to get 12 X squared. Plus 36 x minus 48. And if we want to identify the critical points, we want to set it to 0, right? It exists for all the domain of F of X. So essentially what we have to do is factor out 12, and we're going to get X squared. Plus 3 x minus 4, which is equal to 0, we can factor further because we have a quadratic trinomial. So this gives us 12 multiplied by x + 4 multiplied by x minus 1, which is equal to 0. And this gives us two possible solutions, right? Our first factor can be equal to 0, or the second factor can be equal to 0 if we treat 12 as the leading coefficient. So essentially, we got -4 and 1 as our possible solutions, right? So, -1, -4, I'm sorry, and 1. Those are the critical X values. Now we're going to identify the second derivative because we want to use the second derivative test. So we're going to differentiate 12 x 2 + 36 x minus 48. Once again, let's apply the power rule. This gives us 24 X plus 36, right? And now, what we want to do is basically evaluate the second derivative at each critical point. So F at -4 gives us 24 multiplied by -4. And then we are adding 36. So we get 96 plus 36, right? So, essentially, even without calculating the value, we can say that it is negative, right? Because the negative term has A larger magnitude than the positive term and now F at 0.1 would be equal to 24 multiplied by 1 + 36, we can see that we have a sum of two positive terms, so it's positive. Now let's recall that if the second derivative is negative, the function is concave down, and it represents a local maximum. And if the second derivative is positive, the function is concave up, and this represents a minimum, right? So we have identified our critical X values and now we can identify. The minimum and maximum points. To do that, we're going to evaluate the function at each critical point, so F at -4 would be equal to 4 multiplied by -4 cubed plus 18 multiplied by -4 square minus 48 multiplied by -4. And this gives us 224. Then we want to evaluate F at 0.1. This gives us 4 multiplied by 1 cubed plus 18 multiplied by 1 squad minus 48 multiplied by 1, and this gives us -26. We know that our minimum is at 1, which is indeed our minimum because it's lower than 224. So we can say that F. Minimum Which means the minimum value of the function F. Now, at which point do we get it? Well, we get it at 0.1 so we're saying that the minimum value of the function F is at 0.1 and that minimum is equal to -26. While the maximum value of max is observed at X equals -4, and that maximum is 224, so those will be our answers to this problem. Thank you for watching.

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.

p(t) = 2t³ + 3t² - 36t

Key Concepts

Critical Points