`f(x) = (x^2)e^(-x)` (a) Use a graph of `f` to estimate the maximum and minimum values. Then find the exact values. (b) Estimate the value of `x` at which `f` increases most rapidly. Then find...

`f(x) = (x^2)e^(-x)` (a) Use a graph of `f` to estimate the maximum and minimum values. Then find the exact values. (b) Estimate the value of `x` at which `f` increases most rapidly. Then find the exact value.

Asked on by enotes

Textbook Question

Chapter 4, 4.3 - Problem 56 - Calculus: Early Transcendentals (7th Edition, James Stewart).
See all solutions for this textbook.

1 Answer | Add Yours

gsarora17's profile pic

gsarora17 | (Level 2) Associate Educator

Posted on

a)  Graph is attached . Minimum value of function is 0 at x=0

`f(x)=x^2e^-x`

`f'(x)=x^2d/dxe^-x+e^-xd/dxx^2`

`f'(x)=x^2(-1e^-x)+e^-x(2x)`

`f'(x)=e^-x(-x^2+2x)`

`f'(x)=-e^-x*x(x-2)`

So the critical numbers can be evaluated for f'(x)=0

Critical numbers are x=0 , x=2

f(0)=0

So the function has Minimum=0 at x=0

Images:
This image has been Flagged as inappropriate Click to unflag
Image (1 of 1)

We’ve answered 318,994 questions. We can answer yours, too.

Ask a question