Applications Of Derivatives Question 108
Question: The maximum value of $ 2x^{3}-24x+107 $ in the interval [-3, 3] is
Options:
A) 75
B) 89
C) 125
D) 139
Show Answer
Answer:
Correct Answer: D
Solution:
Let $ f(x)=2x^{3}-24x+107 $
At $ x=-3,\ f(-3)=2{{(-3)}^{3}}-24(-3)+107=125 $
At $ x=3,\ \ f(3)=2{{(3)}^{3}}-24(3)+107=89 $
For maxima or minima, $ {f}’,(x)=6x^{2}-24=0 $
$ \Rightarrow x=2,\ \ -2 $ So at $ x=2,\ f(2)=2{{(2)}^{3}}-24(2)+107=75 $ at $ x=-2,\ \ f(-2)=2{{(-2)}^{3}}-24(-2)+107=139 $
Thus the maximum value of the given function in [- 3, 3] is 139.
BETA
