# rule of differentiation: mean value theorem

• Oct 22nd 2008, 07:02 AM
collegestudent321
rule of differentiation: mean value theorem
hi, guys. i'm having trouble solving this problem. i have a feeling i am making it more difficult than it really is. I really need help because the hw is due today! please help!

show that (1+x)^(1/2) < 1 + (1/2)x if x > 0
• Oct 22nd 2008, 07:55 AM
collegestudent321
anyone????
• Oct 22nd 2008, 07:55 AM
Tony2710
Hi,

The first thing I'd do is get rid of the square root by squaring both sides of the equation. then cancel out some x's and units, and you should be just about there with the proof.
• Oct 22nd 2008, 07:59 AM
collegestudent321
yea, i get how to do it algebraically, but it says to use the mean value theorem. that's what has me confused.
• Oct 22nd 2008, 08:06 AM
Plato
Suppose that $x > 0$. Because $f(x) = \sqrt {1 + x}$ has a derivative, by the mean value theorem
$\left( {\exists c \in \left( {0,x} \right)} \right)\left[ {\frac{1}{{2\sqrt {1 + c} }} = \frac{{\sqrt {1 + x} - 1}}{x}} \right]$.
But we know that $\frac{1}{{2\sqrt {1 + c} }} < \frac{1}{2}$.
Now you put all that together to finish it all off.
• Oct 22nd 2008, 08:14 AM
collegestudent321
wow, ok, i see. thank you soooo much!