Let f(x)=x3-4x2+x+1. Use the tools of calculus to find the absolute minimum and the absolute maximum on the interval [0,3].

Question

Let f(x)=x³-4x²+x+1. Use the tools of calculus to find the absolute minimum and the absolute maximum on the interval [0,3].

Answer 1

f(x) = x³ - 4x² + x + 1, 0 ≤ x ≤ 3

Since f(x) is continuous on the closed interval [0,3], f(x) has maximum and minimum values that occur either at a critical point in the interval (0,3) or at an endpoint.

Find critical points: f'(x) = 3x² - 8x + 1 = 0

x = [8 ± √52] / 6 ≈ 2.535 or 0.131

Evaluate f(x) at each critical point and at each endpoint:

f(0) = 1

f(3) = -5

f(0.131) = 1.065 (absolute max value)

f(2.535) = -5.879 (absolute minimum value)

Answer 2

The absolute maximum of f(x)) on [0,3] is 1 at x=0, and the absolute minimum is approximately −5.85 at x=2.53.