# Find the maximum and minimum values of:

1. Aug 1, 2010

### Nawz

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

Find the maximum and minimum values of

f(x)=x^4e^-x over [0,10]

2. Relevant equations

3. The attempt at a solution

From what I remember is you set the f' equal to 0. And that will give you two more points to test. and then you put the 0 and 10 number into the equation to find the values.

When I put 0 or 10 in i get 0 for both.

f'= (4x^3)(e^-x)+(x^4)(-e^-x)

If i set that equal to 0; I really do not know how to find the what x is.

Last edited: Aug 1, 2010
2. Aug 1, 2010

### Char. Limit

Well, first, let's write this out...

$$f'(x) = 4 x^3 e^{-x} - x^4 e^{-x} = 0$$

First, I'd recommend factoring.

3. Aug 1, 2010

### Nawz

okay i see that now. so you get something like:

(x3)(e-x)[4-x]

so then

(x3)(e-x)=0

and (4-x)=0

so you get x=4

and then i think the other one is ln0/-1 which is undefined if you set it to 0.

4. Aug 1, 2010

### Nawz

Yep, okay i got it. Maximum is 4.6888 at x=4. and Min is 0 at x=0

I don't know if (x3)(e^-x)=0 gives you ln0/-1 but either way 4 is the value I was looking for.

Thank You

Last edited: Aug 2, 2010
5. Aug 1, 2010

### Char. Limit

Hmm... I'm not sure where you're getting -log(0)... You know that if $x^3 e^{-x} = 0$, then either x^3=0 or e^(-x)=0, and only one of those has a solution.

You've found a critical point at x=4, but there's one more.

6. Aug 2, 2010

### Nawz

If you set x^3=0 You get x=0?
If you set e^-x=0 You multiply both by (ln) so -x =ln(0) divide both by -1. That is what I thought? ln(0) / -1

7. Aug 2, 2010

Yes, but since $ln(0)=-\infty[/tex], what you essentially said is that e^(-x)=0 for an x outside the standard real (and complex, incidentally) number system. So, it doesn't count as a root. Also, your domain was [0,10]. Negative infinity is outside that domain. 8. Aug 2, 2010 ### HallsofIvy I would be inclined to write [itex]fsingle-quote(x) = 4 x^3 e^{-x} - x^4 e^{-x} = 0$ as $4x^3e^{-x}= x^4e^{-x}$. Since $e^{-x}$ is never 0, we can divide both sides of the equation by it and get $4x^3= x^4$. If x is not 0, we can divide both sides by x to get x= 4.

That tells you the two roots of the equation.