Applications Of Derivatives Question 364
Question: The minimum value of function $ f(x)=3x^{4}-8x^{3}+12x^{2}-48x+25 $ on
[0, 3] is equal to [Pb. CET 2004]
Options:
A) 25
B) - 39
C) - 25
D) 39
Show Answer
Answer:
Correct Answer: B
Solution:
$ f(x)=3x^{4}-8x^{3}+12x^{2}-48x+25 $ $ f’(x)=12x^{3}-24x^{2}+24x-48 $
$ =12[x^{3}-2x^{2}+2x-4] $
$ =12[(x-2)(x^{2}+2)] $
For maximum and minimum value of the function $ f’(x)=0 $
therefore $ x=2 $ . Now $ {f}’’(x)=12[3x^{2}-4x+2] $ $ {f}’’(2)=12,[12-8+2]=72>0 $
Hence the function is minimum at $ x=2 $
Minimum value of $ f(x) $ on [0, 3] $ =\min { f(0),,f(2),,f(3) } $
$ =\min { 25,,-39,,16 } $
$ =-39 $ .
BETA
