Applications Of Derivatives Question 291
Question: The maximum value of function $ x^{3}-12x^{2}+36x+ $ 17 in the interval [1, 10] is
Options:
A) 17
B) 177
C) 77
D) None of these
Show Answer
Answer:
Correct Answer: B
Solution:
Let $ f(x)=x^{3}-12x^{2}+36x+17 $
$ \therefore f’(x)=3x^{2}-24x+36=0 $ at $ x=2,,6 $ Again $ {f}’’(x)=6x-24 $ is $ -ve $ at $ x=2 $ So that $ f(6)=17,\ \ f(2)=49 $ At the end points $ =f(1)=42,f(10)=177 $ So that $ f(x) $ has its maximum value as 177.
BETA
