Applications Of Derivatives Question 194
Question: The minimum value of the function $ y=2x^{3}-21x^{2}+36x-20 $ is
[MP PET 1999]
Options:
A) -128
B) -126
C) -120
D) None of these
Show Answer
Answer:
Correct Answer: A
Solution:
Given, $ f(x)=2x^{3}-21x^{2}+36x-20 $
$ f’(x)=6x^{2}-42x+36 $ Put $ f’(x)=0 $
therefore $ 6x^{2}-42x+36=0 $
therefore $ x^{2}-7x+6=0 $
therefore $ x^{2}-6x-x+6=0 $
therefore $ (x-1)(x-6)=0 $
therefore $ x=1,,6 $ Now, $ {f}’’,(x)=12x-42 $
$ {f}’’,(1)=-30=-ve $ and $ {f}’’,(6)=30=+ve $
Hence $ x=6 $ is the point of minima Minimum value = $ f(6)=2{{(6)}^{3}}-21{{(6)}^{2}}+36\times 6-20 $
$ f(6)=-128 $ .
BETA
