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For what values of the numbers $ a $ and $ b $ does the function$$ f(x) = axe^{br^2} $$have the maximum value $ f(2) = 1 $?

$a=\frac{\sqrt{c}}{2}$

02:22

Wen Z.

Calculus 1 / AB

Calculus 2 / BC

Chapter 4

Applications of Differentiation

Section 3

How Derivatives Affect the Shape of a Graph

Derivatives

Differentiation

Volume

Baylor University

University of Nottingham

Boston College

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So we're given this function which is equal to a. X times E. To the b times X square. So what we're gonna do is we're gonna find the values of A. And B. Such that F. Of two which is equal to one is a maximum. So we know that if f. of two is a one or is equal to one is a maximum, that means that the derivative at two is going to be equal to zero. So the first thing that I'm going to do is find the derivative of F. Of X. And we're gonna have to use the product rule and the chain rule to find this. So the product rule is the derivative of the first times a second. So is just a constant. The root of X is one. So we just have times a second, which is E. To the B X squared. And then we have plus the first times the derivative of the second. And the derivative of E to the B X squared is E. To the B X squared times the derivative of whatever's in this exponent, which would be to be X. And now I'm just going to simplify this a little bit and we can actually take out A times E to the B X squared out of this. In factor that first, that's what I'm gonna do. And then we're left with one plus we took an A. Either be X squared out. So we're left with X times to be X. So this would be plus to be X squared. And now we know that if the derivative at two is equal to zero so we can plug into and set this equal to zero. So you can say zero is equal to eight times E. To the two squared is four, so four B Plus or times one plus to be times Um two squared would be eight b. And this is equal to zero. So now we have two factors that are multiplied by each other equal to zero. So we can set each of these factors equal to zero and solve for our A. S. And B's. Well if we solve for a in this first part since each of the four B can never be zero, we know that this be this be part is never going to make this um eight times either for be equal to zero. So only a equaling zero would actually give us zero in this factor. But if we look at our original equation, if A is equal to zero then our whole equation fx is just equal to zero. So we don't want a equaling zero. So this is not what we want. So we're going to find a um we're gonna find what be actually equals and that will be um What will use. So if we -1 over we get eight, B is equal to negative one, then B is equal to negative 1/8. So B is equal to negative 1/8. And that's what we want to use. So now we've got to make sure that this is actually a maximum. So what we're gonna want to do is plug in a value into our derivative that is less than two. Let's plug in. Let's just say zero. And we're gonna make be equal to negative 18 since we know B is equal to negative 1/8. So if we do that we are original derivative was write that down here was A. E. To the B. X squared. Which we now know B is negative 1/8. So it should be negative X squared over eight times one plus. And it was to be X squared. So this would be negative 1/4 X squared. And now if you plug in a value that's less than two. So let's just say this is less than two as plug in zero. We would have eight times each of the zeroth which just would be A. And then multiplied by one plus zero. So this would just be equal to A. So for our for this to be a maximum we would need aid to be greater than zero since we want to be greater than zero so that are derivative F prime of X will be greater than zero at X is equal At X being less than two I guess. Which is what we want. Since we want to go from positive to negative in order to have an actual maximum. So now if we look at values of X being greater than two, let's just say three we would have eight times E to the negative 9/8. So I'm actually going to Just write this out so this is equal to eight times each. The -9 eight Multiplied by one plus negative 1/4 times three squared would be negative 9/4. If we simplify this this would be a over e. to the 9/8 multiplied by one plus 9/4 or one minus 9/4 would be negative 5/4. So again we would have to have a positive A value in order to have a negative value um for our derivative for values of X being greater than two. So again we want a being greater than zero which is good if we wanted a to be less than zero. For this to be um a maximum than we would have conflicting conditions and we wouldn't actually be able to have a maximum with B being equal to negative 1/8. But it is good that a. Being greater than zero is the condition for both of these. So now the only thing we have to do is actually plug in this value of two into our original equation which was A. X. Two times E. To the B. X. Squared. So this is to the negative X squared over eight. And this is actually just F. Of X. Not F two. And now if we plug in F of two when we know F of two has to equal one. So you can say one is equal to a times two times E. To the negative two squared is 4/8 would be negative one half. So now we can multiply by this E. To the one half on both sides to get rid of this E. So we get E to the one half is equal to to A. So A is equal to E. To the one half divided by two. So we've now found the values for A. And B. Where A is equal to the square root of E, divided by two, or E. To the one half divided by two, and B is equal to negative 1/8.

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