Absolute Max/Min of y = x^2 * e^-x on [-1,3]

[BuBbLeS01]

Homework Statement

Find the absolute max/min on [-1,3] for y = x^2 * e^-x

Homework Equations

The Attempt at a Solution

F'(y) = (2x)(e^-x) + (x^2)(e^-x)
= xe^-x(2 + x)
x = 0, -2 (Critical Points)
F(-2) = 29.56 MAX (CP)
F(-1) = 10.87
F(0) = 4
F(3) = 0.199 MIN (Right Endpoint)

Can someone please confirm if I am doing this correctly?

 

[Dick]

The derivative of e^(-x) is -e^(-x), isn't it? Chain rule.

 

[BuBbLeS01]

The derivative of e^(-x) is -e^(-x), isn't it? Chain rule.

Ohhhh yeaaaa...thats right. But now how do I reduce it so I can get my critical points?

 

[BuBbLeS01]

F'(y) = (2x)(e^-x) + (x^2)(-e^-x)
= xe^-x(-2 + x)
Critical Points = x = 0, 2

 

[Dick]

Ok, so far. Continue.

 

[BuBbLeS01]

Ok, so far. Continue.

F'(y) = (2x)(e^-x) + (x^2)(e^-x)
= xe^-x(2 + x)
x = 0, -2 (Critical Points)

Plug CP's and endpoints into original F(x):

F(-2) = 11.39 MAX (CP)
F(-1) = 3.72 (ENDPOINT)
F(0) = 1 MIN (CP)
F(3) = 9.05 (ENDPOINT)

 

[Dick]

Why did -2 turn back into 2 again??

 

[BuBbLeS01]

oops my mistake

 

[Dick]

oops my mistake

Not the x=-2 critical point, the 2 in your factorization. Why are you making so many mistakes? The critical points as you said once upon a time are x=0 and x=2. And how can f(0) be 1?? Get a cup of coffee and a clean sheet of paper and start this all over again. Don't cut and paste from previous posts.

 

[BuBbLeS01]

F'(y) = (2x)(e^-x) + (x^2)(-e^-x)
= e^-x(2x - x^2)
= xe^-x(2 - x)
x = 0, 2 (Critical Points)

I Plugged CP's and endpoints into original F(x):

F(2) = 0.54 (CP)
F(-1) = 2.72 MAX (ENDPOINT)
F(0) = 0 MIN (CP)
F(3) = 0.45 (ENDPOINT)

Max = -1
Min = 0

 

[HallsofIvy]

The maximum value is at x= -1. The maximum value itself is e, not "2.7". (No reason to do the extra work-of using a calculator- to get an approximation when you already have an exact value.) The minimum value is at x= 0 and the minimum value is 0.