Help with Y'=0 wanted

1. Aug 30, 2012

jakobs

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

I'm gonna solve the equation Y'=0 if Y=x*e-0,4x

2. Relevant equations

3. The attempt at a solution

I can come as far as to Y'=(1-0,4x)*e-0,4x

Where do I go from here?
Can i just write (1-0,4x)*e-0,4x=0 ?

I can solve the easier kinds of these equations, but this one is the hardest of the ones that I have, and I suspect that something like this will show up on a test in the future, so it would be good if I can solve it.

Anyone can help me in the right direction?

2. Aug 30, 2012

drawar

Since $${e^a}$$ is always > 0 for every value of a, then assume your computed y' is correct, you only need to solve the equation: 1-0.4x = 0, as simple as that.

3. Aug 30, 2012

Staff: Mentor

Yes. Now, either 1 - 0.4x = 0 or e-.4x = 0.

Note that e-.4x ≠ 0 for any real x.

Edit: Didn't notice that drawar said essentially the same thing.

4. Aug 30, 2012

jakobs

So I only have to solve 1-0,4x=0 and that will be the whole answer for the whole equation?

5. Aug 30, 2012

CAF123

Well, the solution to 1-0.4x =0 solves Y'=0.

6. Aug 30, 2012

jakobs

I don't understand the whole thing

Can somebody show all steps to solving this one?
I really need to learn it.

7. Aug 30, 2012

Staff: Mentor

The problem, apparently, is to find the x value(s) for which f'(x) = 0, where f(x) = xe-.4x. (Changed from your notation of y(x) to f(x).)

You found f'(x) = (1 - 0.4x)e-.4x

If f'(x) = 0, then (1 - 0.4x)e-.4x.

For what x is f'(x) = 0?

8. Aug 30, 2012

jakobs

According to some earlier posts the solution to 1-0.4x =0 solves Y'=0

If x is 2.5 then it will be 0.

But what should I do with e-0,4x?

9. Aug 30, 2012

Staff: Mentor

Nothing. As already mentioned, e-0.4x > 0 for all real x.

10. Aug 30, 2012

jakobs

Okay, so the answer to the problem is:
y=x*e-0,4x
y'=(1-0,4x)*e-0,4x

y'=(1-0,4x=0
x=2.5

e-0,4x is always >0

And this classifies as the correct answer?

11. Aug 30, 2012

Staff: Mentor

You have a lot of cruft in there that is unnecessary. Here is all you need to say:

If y = x*e-0.4x, then y' = 0 when x = 2.5.