**Find the absolute extrema of the function on the closed interval.**

**f(x) = sinx - cosx; [0,π]**

**Answer:**

f’(x) = cos(x) + sin(x)

0 = cos(x) + sin(x)

-sin(x) = cos(x)

x = 3pi/4

f’(pi/2) = 1

f’(pi) = -1

using the first derivative test you can find out it changes from a positive slope to a negative slope which means it is a relative max.

Relative Max at the point (3pi/4, sqrt2)