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) = 6x² - x³

Accepted Answer

Hi everyone, let's take a look at this practice problem. This problem says to locate the critical points of H of X, which is equal to XQ minus 6 X2 + 9 X, and use the second derivative test to identify whether these points are local maxima, minima, or neither. So the question asks us to first uh locate the critical points of our function H of X. So to do this, we'll need to take a derivative with respect to X of our function. So we'll need to calculate H of X. And this is going to be equal to the derivative with respect to X. Of the quantity of X cubed minus 6 X squad. Plus 9 X. And so taking this derivative, this is going to be equal to 3 X 2 minus 12 X. Plus 9. So we call for a critical point. We either need to have the first derivative equal to 0, or the first derivative to not exist. Now here we have a polynomial as our first derivative, and polynomials are defined everywhere on the real line. So in order to find the critical points, we need to just set this expression equal to zero and solve for X. So doing so, we will have 3 X squad minus 12 X plus 9 is equal to 0. I can divide both sides of this equation by 3 to simplify the expression, and we will get X2 minus 4 X. Plus 3 is equal to 0, and we can actually factor the left hand side of this equation. So we'll have the quantity of X minus 1 in quantity multiplied by the quantity of X minus 3 in quantity, and that is equal to 0. So now I just need to set each factor equal to 0 and solve for X. So, for the first factor, we'll have X minus 1 equal to 0. Solving for X by adding 1 to both sides, we'll get X is equal to 1. For the other factor, we will set X minus 3 equal to 0. And we need to solve this now for X by adding 3 to both sides of this equation, and we will get X is equal to 3. So these are my two critical points. And so now I need to use the second derivative test to determine whether these points are maxima, minima, or neither. So, we first need to begin by calculating the second derivative of our function H of X. So we'll need to calculate H of X. And this is going to be equal to the derivative with respect to X. Of the quantity, and here we'll use our first derivative, which is 3 x 2. -12 X. Plus 9. So taking this derivative, this is going to be equal to 6 X minus 12. So now we just need to substitute in the value for X for each of our critical points. So, the first one we're going to look at, that's when X is equal to 1. We'll have H double of 1. is equal to 6 multiplied by 1 minus 12, and this is going to be equal to minus 6 when we evaluate that expression. So here we see that our second derivative here is negative. And so that means that X equal to 1 is actually a maximum. So, X equals 1 is going to be a local maximum. So now we're gonna have to apply something similar for X equal to 3. So we'll have a prime of 3. is equal to 6 multiplied by 3 minus 12, and evaluating this expression, this is going to be equal to 6. So here, our second derivative is actually positive. So this means that we actually have a local minimum for our critical point. So at X equal to 3. We have a local minimum. So, for our final answer, we see that we have a local maximum at X equal to 1, and we have a local minimum at X equal to 3. So those are the answers for this problem. 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) = 6x² - x³

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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) = 6x² - x³