Use the Mean Value Theorem to prove that sqrt(1+x) < 1 + x/2 for all x > 0. Generalize this result to the function (1+x)^r, where r < 1 is a positive rational number.

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- Oct 11th 2009, 10:14 AMfriday616mean value theorem
Use the Mean Value Theorem to prove that sqrt(1+x) < 1 + x/2 for all x > 0. Generalize this result to the function (1+x)^r, where r < 1 is a positive rational number.

Thanks! - Oct 11th 2009, 10:40 AMredsoxfan325
Using the MVT, $\displaystyle \frac{f(x)-f(0)}{x-0}=\frac{\sqrt{1+x}-1}{x}=f'(t)$ for some $\displaystyle t\in[0,x]$.

$\displaystyle f'(t)=\frac{1}{2\sqrt{1+x}}$. For $\displaystyle x>0$, $\displaystyle f'(t)<\frac{1}{2}$.

Thus,

$\displaystyle \frac{\sqrt{1+x}-1}{x}<\frac{1}{2}\implies\sqrt{1+x}<1+\frac{x}{2}$ for $\displaystyle x>0$

as desired.

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Using the same method, see if you can generalize that result.