Determine the absolute maximum and the absolute minimum value of th

Question

Determine the absolute maximum and the absolute minimum value of the function

f (x) = sin^2 x X-SQRT2cos x

on the interval (0,3pi/2)

Answer 1

Assuming that the function is f(x) = sin²x - √2cosx and that the interval is [0, 3π/2], the function is continuous on the interval. So, absolute max and absolute min on the interval are guaranteed to exist. The absolute extrema can occur only at critical points in the interval or at endpoints.

Find critical points in the interval:

f'(x) = 2sinxcosx + √2sinx = 0

sinx(2cosx + √2) = 0

sinx = 0 or cosx = -√2/2

sinx = 0 when x = 0 and π

cosx = -√2/2 when x = 3π/4 or 5π/4

Evaluate f at each critical point in the interval and at each endpoint:

f(0) = -√2

f(3π/4) = 1/2 + 1/2 = 1

f(π) = √2

f(5π/4) = 1/2 + 1/2 = 1

f(3π/2) = 1

Abs max value = 1 and abs min value = -√2.

Answer 2

When finding maxima and minima, you need to look for critical points (include boundary points, too). Do you know how to find those?