Originally Posted by

**physics4life** how would i find something like:

$\displaystyle max[\frac{x}{2} + cos(x)]$ where $\displaystyle x\epsilon [0,\pi]$ and ...

$\displaystyle max[\frac{x}{2} + cos(x)]$ where $\displaystyle x\epsilon [-\pi,0]$

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**here's what i did for **$\displaystyle max[\frac{x}{2} + cos(x)]$ where $\displaystyle x\epsilon [0,\pi]$:

first we need critical points: so f'(x) = 0

so 1/2 - sin(x) = 0 , so $\displaystyle x=\frac{\pi}{6} and \frac{5}{6}\pi$

maxf(x) = max{f(0), f(pi/6), f(5/6pi), f(pi)}

f(0) = 1/2(0) + cos(0) = 1

f(pi/6) = 1/2(pi/6) + cos(pi/6) = pi/12 + $\displaystyle \sqrt\frac{3}{2}$

f(5/6pi) = 5/12pi + cos(5/6pi) = 5/12pi - $\displaystyle \sqrt\frac{3}{2}$

f(pi) = pi/2 - 1

so max is at f(pi/6) .. i.e pi/12 + $\displaystyle \sqrt\frac{3}{2}$

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now for $\displaystyle max[\frac{x}{2} + cos(x)]$ where $\displaystyle x\epsilon [-\pi,0]$

critical points are again $\displaystyle x=\frac{\pi}{6} and \frac{5}{6}\pi$, but these both lie outside the range: $\displaystyle [-\pi,0]$ so we only take into consideration -pi and 0:

so f(-pi) = -pi/2 - 1

f(0) = 1

so the max is at point f(0).. i.e 1..

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are these both correct? although seeming a bit obvious.. or do i need to show the final answer differently?