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) = eˣ(x - 2)²

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 E rates to the X, multiplied by the quantity of 2 X minus 3 in quantity squared, and use the second derivative test to identify whether these points are local maximum, minima, or neither. Now, the first part of this question wants us to find the critical points of our function F of X. So, in order to find the critical points, we need to first take a derivative of our function with respect to X. So we'll need to calculate F of X. That is going to be equal to the derivative. With respect to X of the quantity of E raised to the power X, multiplied by the quantity of 2 X minus 3 in quantity squared. So taking this derivative, this is going to be equal to, and here we're going to have to use our product rule. So we call for the product rule that we will have the derivative with respect to X of the first terminal product, so we'll have the derivative with respect to X of E to the X, which is going to be erased the X. Then that gets multiplied by the second term in our product, which is going to be the quantity of 2 X minus 3 in quantity squared. Then we'll have plus. The first term in our product, which is E X, multiplied by the derivative of the second term in our product. So that will be the derivative with respect to x of the quantity of 2 X minus 3 in quantity squared. So first I'll apply the power rule and I will get 2 multiplied by the quantity of 2 X minus 3 in quantity. Then we'll have to apply the chain rule, so I'll need to take the derivative with respect to X of the quantity of 2X minus 3, which is just 2. So I'll multiply this by 2. So, on the right hand side here, I can factor out an E to the X. So this is going to be equal to E to the X. Multiplied by the quantity, and here we multiply everything else out on the right hand side. And so I will get 4 X squad. Minus 12 X. Plus 9 Plus 8 X. -12. So simplifying this expression, this is going to be equal to E to the X, multiplied by the quantity of 4 X squared. Minus or X. -3 in quantity. So now we need to use this to find our critical points. So we call for a critical point to exist. We need the first derivative to either not exist, or we need the first derivative to be equal to zero. Now here we see that our first derivative is a composition of an exponential and a polynomial function. And since each one of those, the exponential and the polynomial, are continuous over the entire red line, that means the composition is also continuous. So that means that our function FX exists everywhere. So in order to find the critical points, we need to set our first derivative here equal to 0. That means that we will have EX multiplied by the quantity of 4 X squad minus 4 X. minus 3 in quantity has to be equal to 0. Now, the exponential, the E to the X can never be zero, so I can divide both sides of the equation by E X. And so that means that the polynomial has to be equal to 0. So we'll have 4 X squared, minus 4 X, minus 3 is equal to 0. Now we can actually factor this quadratic, and this factors into the quantity of 2 X plus 1 in quantity multiplied by the quantity of 2 X minus 3 in quantity, that is equal to 0. So now we'll set each one of these factors equal to 0 and solve for X. So for the first factor, we will have 2 X plus 1 is equal to 0, and solving this for X we'll get X is equal to minus 1 divided by 2. Now for the second factor, we will have 2 X minus 3 is equal to 0, and solving for X, we will get X is equal to 3 divided by 2. So these are our two critical points. We have one at X equal to minus 1 divided by 2 and another critical point at X equals to 3 divided by 2. So now we need to use our second derivative test to see whether these are local, maxima, minima, or neither. So, the first thing we need to do is calculate our second derivative. So we will need to look at F of X. This is going to be equal to the derivative with respect to X of the quantity of eat the X, multiplied by the quantity of 4 X squared minus 4 X minus 3 in quantity. So taking this derivative, this is going to be equal to, and here again, we have to apply our product rule. We'll have the derivative of our first term, which is going to be E to the X, multiplied by the second term, which is going to be 4 X squad, minus 4 X minus 3 in quantity. Then we'll have plus the first term, which is e to the X, multiplied by the derivative of the second term, which is going to be the quantity of 8 X minus 4 in quantity. Again, we can factor out an E X out of both terms. This is going to be equal toe to the X. Multiplied by the quantity, and here we'll just collect like terms. So we'll have 4 X squared. Less or X minus 7 in quantity. So now to use our second derivative test, we will substitute in the value for X for each of our critical points and see whether our second derivative here is positive or negative. So looking at X equals to minus 1 divided by 2, we will have F of minus 1 divided by 2 is equal toe raised to the quantity of minus 1 divided by 2 in quantity, multiplied by the quantity of 4, multiplied by the quantity of -1 divided by 2 in quantity squared, plus, 4 multiplied by minus 1 divided by 2. -7. Evaluating this expression, this is going to be equal to minus 8, multiplied by E rates of the quantity of minus 1 divided by 2 in quantity. And here we can see that. Our value for F at -1 divided by 2 is actually negative. And since the second derivative here is negative, that means that by the second derivative test that our X equal to -1/ critical point is a local maximum. So now we're going to have to repeat the same process, but for X equal to 3 divided by 2. Looking at F double prime. Of 3 divided by 2. This is going to be equal to. E raised to the quantity of 3 divided by 2 in quantity, multiplied by the quantity of 4, multiplied by the quantity of 3 divided by 2 in quantity squared, plus 4 multiplied by 3 divided by 2. 7 in quantity. Evaluating this expression, this is going to be equal to 8, multiplied by E raised to the quantity of 3 divided by 2. So here we see that F of 3 divided by 2 is actually a positive quantity. And so since our second derivative here is positive at the critical point, that means that we have a local minimum. So now we've identified what type of critical point we have. And our final answer here in this case is going to be, we have a local maximum at X equal to minus 1 divided by 2, and we have a local minimum at X equal to 3 divided by 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) = eˣ(x - 2)²

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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) = eˣ(x - 2)²