Extrema of a complicated function.

[coverticus]

Homework Statement

Find the absolute maximum and minimum values of f(x) = e^{}-x - e^{}-2x on [0,1].

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

Homework Equations

The Attempt at a Solution

f '(x) = -e^-x + 2e^-2x
0 = -e^-x + 2e^-2x
e^-x = 2e^-2x
ln(e^-x) = ln(2e^-2x)
-x = ?

 

[neutrino]

ln(2e^-2x) - that's a product of two terms, right?

Another way to do this would be to rewrite the equation (0 = -e^-x + 2e^-2x) as

2\left(e^{-x}\right)^2 - e^{-x} = 0

 

[coverticus]

so you can factor out the e^-x and your left with:
e^-x(2(1)^2-1) = 0, correct?

 

[neutrino]

so you can factor out the e^-x and your left with:
e^-x(2(1)^2-1) = 0, correct?

No. If it makes any easier, substitute y in the place of e^(-x), so that

(2y^2 - y) = 0 becomes y(2y-1) = 0.

 

[Pellefant]

ln(e^-x) = ln(2e^-2x)

I am not sure that he know the rules: ln(A*B)=ln(A)+ln(B)

Also ln is the inverse of e, which means ln(e^C)=C

 

[coverticus]

ok, so after substituting y for e^-x i saw that f has two critical points, on at x=0, and the other where e^-x=(1/2) . But how do you figure out x?

 

[neutrino]

on at x=0,

y=0, actually. e^(-x) = 0.

and the other where e^-x=(1/2) . But how do you figure out x?

Use the log function now.

 

[coverticus]

ok so you take ln(e^-x) = ln(1/2)? What about the other critical number, you can't take ln(0) so how do you solve that function? Or is it a critical number because it's not defined? This is very confusing to me.

 

[coverticus]

what I think I have it. ln(e^-x) = -x, so -x = ln(1/2), or x = -ln(1/2). But what about the zero?

 

[neutrino]

x = -ln(1/2).

Or
x = -ln(1/2) = -(ln(1)-ln(2)) = ln(2) (It just looks better imo. :) )

But what about the zero?

Can you find a real x such that e^(-x) = 0 let alone one within [0,1]?