Absolute maxima and minima Determine the location and value of...

Question

Absolute maxima and minima Determine the location and value of the absolute extreme values of ƒ on the given interval, if they exist.

ƒ(x) = x³e⁻ˣ on [-1,5]

Accepted Answer

Welcome back, everyone. In this problem, we want to find the absolute maximum and the minimum values of the function H on the interval open bracket -1, 3 close bracket. The function is H X equals 5 X cubed multiplied by e-2 X. A says the absolute maximum is -135 divided by 8 E -3 at X equals -1, and the absolute minimum is 5 E 2 at X equals 3 halves. B says the maximum is -135 E 3 divided by 8 at X equals 3 halves, and the minimum is 5 E 2 at X equals -1. CE says the maximum is 135 E to the 3 divided by 8 at X equals 3 halves, while the absolute minimum is -5 E 2 at X equals -1, and the D says it's 15, the maximum is 135 E to the 3 divided by 8 at X equals -1, and the minimum is -5 E squared at X equals 3 halves. Now, how can we find the absolute maximum and minimum values? Well, first, let's make sure we understand our function, OK? Now, function H X equals 5 X cubed multiplied by E negative 2X is a product of the polynomial 5 X cubed and an exponential function E to the negative 2X. That means it's defined and continuous for all real numbers, so it must be well behaved on the interval -1 to 3 inclusive. Now to find the critical points, OK, we know that at the critical point. At the critical point. Which is where our maximum and minimum points may be, then the derivative of H of X is going to be zero. So if we can differentiate H of X and set it to zero, we'll be able to solve for the X values of our critical points. Then we can test whether those and the end or we can test which of those and the end points are going to be the absolute maximum and minimum values or extremes. So first, let's differentiate H of X. Now, if, if we look at our function here, notice that it is a product. So to differentiate it, we can apply the product rule. Recall that by the product rule if we have a function Y that's equal to UV where U and V are functions of X, then the derivative of Y is going to be equal to U V plus UV where U and V are the derivatives of U and V respectively. So if we can let one term be you and one and the other be V, then we can differentiate both and then apply them to the product to. So in this case, if we let you be equal to 5x cubed. And V be equal to E to the negative 2 X, OK. Then that tells us that the derivative of you. If we differentiate 5 X cubed, there is going to be 15 X squared and the derivative of V is going to be equal to -2 negative2X. Thus this tells us OK, let me just clean this up here. This tells us that the derivative of our function H of X. Is going to be equal to UV that's 15 X squared multiplied by E the negative to X. Plus UV that's 5 x cubed multiplied by -2 E to the negative 2 X. Now, if we look at our terms here, notice that we can factor out 5 X squared and E to the negative 2X. And when we do that, if we factor out 5 X squared e to the negative2X, for our first term, it will be 3, and from our second term, it leaves us with 2 X. Thus, this is the derivative of H of X. No, we can set this to 0 to find the critical points. So at 5 X squared multiplied by E to the negative 2X multiplied by 3 minus 2 X equals 0, OK. Then, if we think about it here, since E 2 X and 5 X squared will be, well, E2X is greater than 0 and 5X2 is greater than or equal to 0 for RX, we can divide through for both of those in our equation. And thus We can be left with X2 multiplied by 3 minus 2 X. We can simplify equation to X2d multiplied by 3 minus 2 X to be equal to 0. And when we do that. OK, then. We should get X to be equal to 0, which makes X equals 0 and 3 minus 2 X to be equal to 0, which makes X equal to 3 halves. Thus, these are the critical points. The critical points of our uh equation or function here, OK? Now that we have the critical points, let me just clean this up a little bit. Now that we have our critical points, we can test whether these and or end points give us the global, or sorry, give us the maximum, the absolute maximum and the minimum values for HF X. OK? So evaluating, let me write that here. Evaluating HF X. At the critical points and endpoints. OK. Then we can find it at X equals -1, H of -1, at X equals 0, H of 0, OK? At X equals 3, H of 3 houses, and at X equals 3, OK? Now at X equals -1, H would be equal to 5 multiplied by -1 cubed, multiplied by E to the negative to raised to the negative 2 multiplied by -1. That's gonna be equal to 5 multiplied by -1 E2, which equals -5 E2, OK? Next, When X equals 0, we're gonna have 5 multiplied by 0 cubed multiplied by E to E to the negative 2 multiplied by 0, and that would work out to 0. When X equals 3 halves, you're gonna have 5 multiplied by 3 halves cubed, multiplied by E to the negative 2. Multiplied by the E to the negative 2 multiplied by 3 halves. And when we were that thought, that is gonna be equal to 135 divided by 8, multiplied by E to the negative 3. And then finally, A X equals 3, H would be equal to 5 multiplied by 3 cubed, multiplied by E-2 multiplied by 3, which would work out to be 135 E to the 6. So now that we have all of the values of H for the X values, which ones are the maximum and which ones are the minimum? Well, notice that. This, when X equals 3 halves, it gives us the greatest possible value for H. So 135 divided by 8 E to the negative 3 is the absolute maximum, OK, at X equals 3 halves. On the other hand, -5 is squared would be our least value, OK. So that means that our minimum. Our minimum, our absolute minimum is -5 squad at X equals -1. Therefore, that tells us then. That C is the correct answer. Thanks a lot for watching everyone. I hope this video helped.

Absolute maxima and minima Determine the location and value of the absolute extreme values of ƒ on the given interval, if they exist.

ƒ(x) = x³e⁻ˣ on [-1,5]

Key Concepts

Critical Points

Endpoints of the Interval

First Derivative Test

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