How do you find the relative extrema of

Question

 f(x)= x^2 - 2cos x?

Answer 1

The function is 

 f(x)= x^2 - 2cos x

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

f'(x)= 2x + 2sinx 

make f' (x) = 0

2x +2sin x =0

x +sinx =0

x = - sinx       <- - - - Valid only when x =0

So, x = 0 is only the extreme of the function.

Now evaluating the original function at this point 

f( x) = 0 - 2 cos( 0 )

f (x) =  -2( 1) = - 2 .

So the function has a relative minimum at x = - 2 .