# Max and min of a function

1. Aug 8, 2007

### physics4life

how would i find something like:

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

$$max[\frac{x}{2} + cos(x)]$$ where $$x\epsilon [-\pi,0]$$
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here's what i did for $$max[\frac{x}{2} + cos(x)]$$ where $$x\epsilon [0,\pi]$$:
first we need critical points: so f'(x) = 0

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

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 + $$\sqrt\frac{3}{2}$$

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

f(pi) = pi/2 - 1

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

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

critical points are again $$x=\frac{\pi}{6} and \frac{5}{6}\pi$$, but these both lie outside the range: $$[-\pi,0]$$ 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?

2. Aug 8, 2007

### daveb

Your results at each of the four points for the first one are a little off, but that could just be latex. cos(pi/6) = sqrt(3)/2, not sqrt(3/2). However, your conclusions are correct.

3. Aug 8, 2007

### smoothman

you're right there. the answers are correct as dave said.