Calculus
posted by Tyler on .
find the absolute extrema of f(x)=sin pi x on [1,2]
Not sure how to do this problem....

I read you equation as
f(x) = sin (πx)
f'(x) = πcos(πx)
for local max/min, f'(x) = 0
cos(πx) = 0
πx = π/2 or πx = π/2 or πx= 3π/2
x = ± 1/2 or x = 3/2 , but our domain is [1,2]
so x = 1/2 or x = 1/2
so
f(1/2) = sin (π/2) = 1
f(1/2) = sin (π/2) = 1
also look at endpoints of domain
f(1) = sin(π) = 0
f(2) = sin(2π) = 0
so the max is 1 and the min is 1