Applications Of Derivatives Question 326
Question: The maximum value of $ f(x)=\frac{x}{4+x+x^{2}} $ on $
[-1,,1] $ is [MP PET 2000]
Options:
A) $ -1/4 $
B) $ -1/3 $
C) $ 1/6 $
D) $ 1/5 $
Show Answer
Answer:
Correct Answer: C
Solution:
$ f(x)=\frac{x}{4+x+x^{2}} $
Differentiate, $ {f}’(x)=\frac{4+x+x^{2}-x(1+2x)}{{{(4+x+x^{2})}^{2}}} $
For maximum $ f’(x)=0 $
therefore $ \frac{4-x^{2}}{{{(4+x+x^{2})}^{2}}}=0 $
therefore $ x=2,,-2 $
Both values of x are out of interval
$ f(-1)=\frac{-1}{4-1+1}=\frac{-1}{4} $ , $ f(1)=\frac{1}{4+1+1}=\frac{1}{6} $ (maximum).
BETA
