# local extrema

• Jun 3rd 2009, 08:29 PM
mzhang
local extrema
Hi, I don't know how to find the local extrema, and I'm not sure if I differentiated correctly. for the first one is it -ln [(1-x)^2]-2? how do I find the critical point by letting it 0 =S would somebody be able to show me full solutions for part and b please?
• Jun 3rd 2009, 09:50 PM
pickslides
part b) $f(x) = cos(2x)+ sin(x)$

$f'(x) = cos(x)- 2sin(2x)$

for extreme value

$0 = cos(x)-2sin(2x)$

$0 = cos(x)-2\times 2sin(x)cos(x)$

$0 = cos(x)-4sin(x)cos(x)$

$0 = cos(x)(1-4sin(x))$

$0 = 1-4sin(x)$

$1 = 4sin(x)$

$sin(x)=\frac{1}{4}$

can you solve from here?