# Using Mean Value Theorem to show solutions ?

• May 30th 2009, 11:27 PM
Khonics89
Using Mean Value Theorem to show solutions ?
Suppose xe^-x=(x-1)^2

Use the Mean value theorem to show that this equation has exactly two solutions...

Well I know the mean value theorem is

f'(c)= f(b)-f(a)/b-a

this is the f'(x)

how do we use the first derivative to show a solutions ??

Once again sorry guys- I'm just first year maths student :)

I'm just doing past papers and like they never taught us how to use mean value theorem but its in the exam ???????????

The notes only go in brief ?
• May 31st 2009, 09:39 AM
CaptainBlack
Quote:

Originally Posted by Khonics89
Suppose xe^-x=(x-1)^2

Use the Mean value theorem to show that this equation has exactly two solutions...

Well I know the mean value theorem is

f'(c)= f(b)-f(a)/b-a

this is the f'(x)

how do we use the first derivative to show a solutions ??

Once again sorry guys- I'm just first year maths student :)

I'm just doing past papers and like they never taught us how to use mean value theorem but its in the exam ???????????

The notes only go in brief ?

Consider:

$f(x)=xe^{-x}-(x-1)^2$

$
f'(x)=(1-x)(e^{-x}+2)
$

So $f'(x)<0$ when $x>1$, and $f'(x)>0$ when $x<1$, so we have $f(x)$ is decreasing when $x>1$ and increasing when $x<0$.

Now $f(1)>0$ and $f(0)=-1$, so by the intermediate value theorem as there is a root of $f(x)=0$ between $0$ and $1$, and there can be no other zeros less than $1$ as $f(x)$ is increasing there.

Same argument applies to $x>1$.

CB
• Jun 28th 2009, 07:39 AM
VonNemo19
I am not making the connection to the mean value theorem here...