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

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SUMMARY

The discussion focuses on solving the differential equation Y' = 0 for the function Y = x * e^(-0.4x). Participants clarify that the derivative Y' can be expressed as (1 - 0.4x) * e^(-0.4x). The critical step is solving the equation 1 - 0.4x = 0, which yields the solution x = 2.5. The exponential term e^(-0.4x) is confirmed to be greater than zero for all real x, thus not contributing to the solution of Y' = 0.

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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?
 
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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.
 
jakobs said:

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.

jakobs said:
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.
 
So I only have to solve 1-0,4x=0 and that will be the whole answer for the whole equation?
 
Well, the solution to 1-0.4x =0 solves Y'=0.
 
I don't understand the whole thing :confused:

Can somebody show all steps to solving this one?
I really need to learn it.
 
jakobs said:
I don't understand the whole thing :confused:

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.
jakobs said:
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?
 
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?
 
jakobs said:
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.
 
  • #10
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
jakobs said:
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.
 

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