Find the absolute extrema of the function on the closed interval. f(x) = sinx - cosx; [0,π]

absolute-extrema
closed-interval

#1

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)