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

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

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

Answer»

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