# Math Help - maximum and minimum of a function

1. ## maximum and minimum of a function

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..

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

2. Originally Posted by 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 $x\epsilon [-\pi,0]$
-------------------------------------------
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 + $\sqrt\frac{3}{2}$

---------------------------
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..

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

-Dan