local extrema

**mzhang** · June 3rd 2009, 08:29 PM

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?

**pickslides** · June 3rd 2009, 09:50 PM

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?

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