Max and min of a function

  • #1
how would i find something like:

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

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

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

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 + [tex]\sqrt\frac{3}{2}[/tex]

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

f(pi) = pi/2 - 1

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

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

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

Answers and Replies

  • #2
537
1
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
39
0
you're right there. the answers are correct as dave said.
 

Related Threads on Max and min of a function

Replies
4
Views
4K
  • Last Post
Replies
7
Views
2K
  • Last Post
Replies
5
Views
6K
Replies
7
Views
13K
  • Last Post
Replies
1
Views
4K
Replies
8
Views
4K
  • Last Post
Replies
5
Views
2K
  • Last Post
Replies
2
Views
2K
Replies
4
Views
893
Replies
3
Views
3K
Top