Master the Equation Y'=0 with Expert Homework Help - Solving Tips and Techniques

[jakobs]

Homework Statement

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

Homework Equations

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?

 

[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.

 

[Mark44]

Homework Statement

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

Homework Equations

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 ?

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

Note that e^-.4x ≠ 0 for any real x.

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?

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

 

[jakobs]

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

 

[CAF123]

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

 

[jakobs]

I don't understand the whole thing

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

 

[Mark44]

I don't understand the whole thing

Can somebody show all steps to solving this one?

No, we won't do your work for you, but we'll help you do it.

I really need to learn it.

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?

 

[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?

 

[Mark44]

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?

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

 

[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?

 

[Mark44]

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?

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.