Question

Find the maximum and minimum values of 2x³ - 24x+107 on the interval [-3, 3].

Answer 1

max. value is 139 at x = −2 and min. value is 89 at x = 3

F’(x) = 6x² - 24 = 0

6(x² - 4) = 0

6(x² - 2²) = 0

6(x -2)(x+2) = 0

X = 2,-2

Now, we shall evaluate the value of f at these points and the end points

F(2) = 2(2)³ - 24(2)+107 = 75

F(-2) = 2(-2)³ - 24(-2)+107 = 139

F(-3) = 2(-3)³ - 24(-3)+107 = 125

F(3) = 2(3)³ - 24(3)+107 = 89