Answer to Question #103929 in Calculus for Martin

Question

Find the absolute maximum and minimum values of f(x)=−(x2−x)23 over the interval [-4, 3].

Answer 1

The given function is,

f(x) = -(x² - x)²³

and the interval is [-4,3].

Since, f(x) is a polynomial, it is continuous everywhere. Therefore, it is continuous on the given interval.

Differentiating f(x) to get the first derivative and to calculate the critical points of f(x) :

f'(x) = d(-(x² - x)²³)/dx

f'(x) = (-23(x² - x)²²) × (2x - 1)

To get the critical points, we make f'(x) equal to 0.

i.e., (-23(x² - x)²²) × (2x-1) = 0

i.e., (x² - x)²² × (2x - 1) = 0

i.e., (x² - x)²² = 0 or 2x - 1 = 0

x² - x = 0 or 2x = 1

x(x - 1) = 0 or x = 1/2

x = 0 or x = 1 or x = 1/2

Therefore, we get three critical points at x = 0, x = 1/2 and x = 1, and all these points lie in the given interval.

Evaluating the function at the critical points and the end points :

f(-4) = -( (-4)² - (-4) )²³ = -(16 + 4)²³ = -(20)²³

f(0) = -(0² - 0)²³ = 0

f(1/2) = -( (1/2)² - (1/2) )²³ = -( (1/4) - (1/2) )²³ = - (-1/4)²³ = - (-1)²³ × (1/4)²³ = (1/4)²³

f(1) = -(1² - 1)²³ = -(1 - 1)²³ = 0

f(3) = -(3² - 3)²³ = -(9 - 3)²³ = -(6)²³

Therefore, we see from the above list that the absolute maximum of f(x) is (1/4)²³ which occurs at x = (1/2) (critical point), and the absolute minimum of f(x) is ( - (20)²³ ) which occurs at x = (-4) (end point).