# Thread: Using Mean Value Theorem to show solutions ?

1. ## 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 ?

2. 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

3. I am not making the connection to the mean value theorem here...