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

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- Oct 22nd 2008, 07:02 AMcollegestudent321rule 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 AMcollegestudent321
anyone????

- Oct 22nd 2008, 07:55 AMTony2710
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 AMcollegestudent321
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 AMPlato
Suppose that $\displaystyle x > 0$. Because $\displaystyle f(x) = \sqrt {1 + x} $ has a derivative, by the mean value theorem

$\displaystyle \left( {\exists c \in \left( {0,x} \right)} \right)\left[ {\frac{1}{{2\sqrt {1 + c} }} = \frac{{\sqrt {1 + x} - 1}}{x}} \right]$.

But we know that $\displaystyle \frac{1}{{2\sqrt {1 + c} }} < \frac{1}{2}$.

Now you put all that together to finish it all off. - Oct 22nd 2008, 08:14 AMcollegestudent321
wow, ok, i see. thank you soooo much!