Find the absolute maxima and minima for f(x) on the interval [a, b].

Question

Find the absolute maxima and minima for f(x) on the interval [a, b].

f(x) = x³ − 3x² − 45x + 8,    [−4, 7]

absolute minimum (x,y) = ( ? )

absolute maximum (x,y) = ( ? )

Answer 1

Absolute max and min will occur on an interval in 3 ways:

1) At local max or min values as found by setting the derivative equal to zero (critical points)

2) At the endpoints of the interval (so at f(a) and/or f(b))

3) At any discontinuities in the interval.

To find the possibilities for 1), i.e., critical points, take the derivative, set it equal to zero, and solve:

f(x) = x³ − 3x² − 45x + 8

f '(x) = 3x² - 6x - 45

3x² - 6x - 45 = 0

3(x² - 2x - 15) = 0

3(x - 5)(x + 3) = 0

x = 5 and x = -3

Both of these are on the interval [−4, 7] so are possibilities.

f(5) = (5)³ − 3(5)² − 45(5) + 8 = -167

f(-3) = (-3)³ − 3(-3)² − 45(-3) + 8 = 89

To find the possibilities for 2), i.e., function values at the endpoints, just plug in the endpoints:

f(-4) = (-4)³ − 3(-4)² − 45(-4) + 8 = 76

f(7) = (7)³ − 3(7)² − 45(7) + 8 = -111

For 3), (discontinuities in the interval), since this is a standard polynomial, it has no discontinuities.

So the absolute max is found at x = -3 and is 89

And the absolute min is found at x = 5 and is -167