# 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:

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

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

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

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

Same argument applies to \$\displaystyle x>1\$.

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