# Homework Help: Extrema of a complicated function.

1. Oct 30, 2007

### coverticus

1. The problem statement, all variables and given/known data
Find the absolute maximum and minimum values of f(x) = e$$^{}-x$$ - e$$^{}-2x$$ on [0,1].

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

2. Relevant equations

3. 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 = ?
1. The problem statement, all variables and given/known data

2. Relevant equations

3. The attempt at a solution

2. Oct 30, 2007

### 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

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

3. Oct 30, 2007

### coverticus

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

4. Oct 30, 2007

### neutrino

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.

5. Oct 30, 2007

### 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

6. Oct 30, 2007

### 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?

7. Oct 30, 2007

### neutrino

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

Use the log function now.

8. Oct 30, 2007

### 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.

9. Oct 30, 2007

### 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?

10. Oct 31, 2007

### neutrino

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

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

Last edited: Oct 31, 2007