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xllx
Oct21-09, 10:37 AM
1. The problem statement, all variables and given/known data

0=x(e^-x)

2. Relevant equations



3. The attempt at a solution

Well as x is a multiple that leaves:
0=e^-x
so does x=0?

Any help at all would be greatly appreciated. Many Thanks.

Mark44
Oct21-09, 11:21 AM
If a*b = 0 then a = 0 or b = 0.
For your equation, either x = 0 or e-x = 0.
Clearly, x = 0 is a solution of your equation. Are there any values of x for which e-x = 0?

xllx
Oct21-09, 12:09 PM
Are there any values of x for which e-x = 0?

Well wouldn't x be infinity in that case?
But this question is referring to co-ordinates of a turning point so surely a turning point cant be infinity?

The original equation was y=(x^2)(e^-x)
Which I differentiated into:
dy/dx=x^2(-e^-x)+(e^-x)2x
Which I factorised into:
dy/dx=x(e^-x)[-x+2]

And as dy/dx=0 for turning points then:
0=x(e^-x)[-x+2]
So x=2 and 0=x(e^-x)

Is this all right?

Thankyou so much.

Mark44
Oct21-09, 01:25 PM
So x = 2 or x = ?

xllx
Oct21-09, 02:11 PM
Is it zero?

Mark44
Oct21-09, 02:34 PM
Yes.

xllx
Oct21-09, 02:38 PM
Thankyou!