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!
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.