# Calculus

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find the absolute extrema of f(x)=sin pi x on [-1,2]

Not sure how to do this problem....

• Calculus -

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

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