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) = 2x² ln x - 11x²

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 x 4 multiplied by the natural log of X minus 4 X 4th, and to use the second derivative test to identify whether these points are local maximum, minima, or neither. So the first part of this problem wants us to find the critical points of F of X. So to find our critical points, we'll first need to find the first derivative. So we need to calculate F of X. And this is going to be equal to the derivative with respect to X of the quantity of X to the 4th, multiplied by the natural log of X. Minus or multiplied by X to the 4th. So taking this derivative, this is going to be equal to. For our first term, we'll need to apply our product rule. The recall for the product rule that we'll need to take the derivative of our first term, which in this case is going to be the derivative of X to the 4th with respect to X, that gives me 4 X cubed, then that gets multiplied by the second term in our product, which is the natural log of X. Then we'll have plus, the first term in our product, which is X to the 4th, multiplied by the derivative of the second term in our product. So we'll have the derivative with respect to X of the natural log of X, and that is 1 divided by X. Then we will have minus the derivative with respect to X of 4 X to the 4th, which is 16 X cubed. So, simplifying this equation, this is going to be equal to. 4 X cubed, multiplied by the natural log of X. Minus 15 Xcu. Now recall that critical points are located whenever our first derivative is equal to 0 or where it is also undefined. Now, our first derivative here is undefined for x less than or equal to 0, since the natural log cannot have a negative argument. But if we look at our original function, F of X, it is also not defined for X less than or equal to 0, since it also has the natural function in it. So therefore, we don't really have any critical points due to our first derivative not being defined. So to find the critical points, we need to set our first derivative equal to 0. So doing so, we will have 4 X cubed, multiplied by the natural log of X. Minus 15, multiplied by X cubed is equal to 0. Now I can factor out an X cubed out of both terms on the left hand side. In doing so, I will have X cubed. Multiplied by the quantity, a 4 multiplied by the natural log of X. Minus 15 in quantity is equal to 0. Now we need to set each one of these factors equal to 0 and to solve for X. So I will have X cubed is equal to 0, so that means that X is equal to 0. And we also have 4 multiplied by the natural log of X. Minus 15 is equal to 0, and solving for X will get X is equal to E raised to the quantity of 15. Divided by 4. Now we already said that our function F of X was not defined for x less than or equal to 0, so that means we can ignore the X equal to 0 result here, since it is not part of the domain of our function. And the only critical point that we actually have is X equal to E raised to the quantity of 15 divided by 4. So now we need to use the second derivative test to determine whether this point is a local maximum or minimum or neither. So, to use the second derivative test, we need to first calculate our second derivatives, so we'll need F of X. This is going to be equal to the derivative with respect to X. Of the quantity Of 4 X cubed multiplied by the natural log of X. Minus 15 cubed in quantity. So taking this derivative, this is going to be equal to. And here again, on the first term, we'll have to use our product rule. So taking the derivative of the first term, we will get 12 x squad, multiplied by the second term, which is the natural log of X. Plus, the first term, which is for X cubed. Multiplied by the derivative of the natural log of X, which is 1 divided by X. And then we'll have minus. 45 X squad for the last term. So simplifying this expression, this is going to be equal to. 12 x squared multiplied by the natural log of X. Minus 41 X squared. So now what we need to do is substitute in the value for a critical point for X and see whether our second derivative here, FX is positive or negative. If it is positive, then that means that we have a local minimum, and if it's negative, that means we have a local maximum. So, for F double prime of E rays of the quantity of 15 divided by 4 in quantity, this is going to be equal to. 12 multiplied by the quantity of E raised to the quantity of 15 divided by 4 in quantity squared multiplied by the natural log of e raised to the quantity of 15 divided by 4 in quantity minus 41 multiplied by the quantity of E raised to the quantity of 15 divided by 4 in quantity squared. So, evaluating this expression, this is going to be equal to. Or multiplied by E rates of the quantity of 15, divided by 2 in quantity. So here we see that our second derivative at our critical point is positive. And so that means that we have a local minimum. Uh, at that point. So, the final answer that we will have for this problem is that we have a local minimum at X equal to E raised to the quantity of 15 divided by 4 in quantity. 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) = 2x² ln x - 11x²

Key Concepts

Critical Points