text: 1. ## 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?
  
  2. 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?