
1 Attachment(s)
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 [(1x)^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?

part b) $\displaystyle f(x) = cos(2x)+ sin(x)$
$\displaystyle f'(x) = cos(x) 2sin(2x)$
for extreme value
$\displaystyle 0 = cos(x)2sin(2x)$
$\displaystyle 0 = cos(x)2\times 2sin(x)cos(x)$
$\displaystyle 0 = cos(x)4sin(x)cos(x)$
$\displaystyle 0 = cos(x)(14sin(x))$
$\displaystyle 0 = 14sin(x)$
$\displaystyle 1 = 4sin(x)$
$\displaystyle sin(x)=\frac{1}{4}$
can you solve from here?