Question

find absolute max and min of f(x)=cosx+sin^2(x) on interval [0,2pi]. Give exact answers, not decimal approx. (Use closed interval method)

Answer 1

f(0) = 1+0 = 1
f(2pi) = 1+0 = 1

f'(x) = -sinx + 2sinx cosx
= sinx(2cosx-1)

So, f'=0 at x=0,pi,2pi and pi/3,5pi/3

f(pi/3) = 1/2 + 3/4 = 5/4
f(pi) = -1+0 = -1
f(5pi/3) = 1/2 + 3/4 = 5/4

Looks like the min is -1 and the max is 5/4

See

http://www.wolframalpha.com/input/?i=cosx%2Bsin^2%28x%29