Question

locate the absolute extrema of the function f(x)=cos(πx) on the closed interval [0,3/4] .

a) absolute max: f(3/4)=cos3π/4, absolute min: f(0)=1
b) absolute max: f(0)=1, absolute min: f(3/4)=cos3π/4
c) absolute max: f(3/4)=cos3π/4, No absolute min
d No absolute max, absolute min: f(0)=1

Answer 1

Answer:

b) absolute max: f(0)=1, absolute min: f(3/4)=cos3π/4

Step-by-step explanation:

Find the derivative.

f'(x) = -π sin(πx)

Set to 0 and solve.

0 = -π sin(πx)

πx = 0 + kπ

x = 0 + k

On the interval [0, 3/4], x = 0 is an extrema (either local min or local max).

Now evaluate the end points.

f(0) = cos(0) = 1

f(3/4) = cos(3π/4) = -√2/2

Absolute max at f(0) = 1, absolute min at f(3/4) = cos(3π/4).