Applications Of Derivatives Question 324
Question: The denominator of a fraction number is greater than 16 of the square of numerator, then least value of the number is
[RPET 2000]
Options:
A) $ -1/4 $
B) $ -1/8 $
C) $ 1/12 $
D) $ 1/16 $
Show Answer
Answer:
Correct Answer: B
Solution:
The function $ f(x)=\frac{x}{x^{2}+16} $
therefore $ {f}’(x)=\frac{(x^{2}+16).1-x.(2x)}{{{(x^{2}+16)}^{2}}} $ = $ \frac{x^{2}+16-2x^{2}}{{{(x^{2}+16)}^{2}}}=\frac{16-x^{2}}{{{(x^{2}+16)}^{2}}} $ –(i) Put $ {f}’(x)=0 $
therefore $ 16-x^{2}=0 $
therefore $ x=4,,-4 $
Again, $ f’’(x)=\frac{{{(x^{2}+16)}^{2}}(-2x)-(16-x^{2})2(x^{2}+16)2x}{{{(x^{2}+16)}^{4}}} $
At $ x=4 $ , $ {f}’’(x)>0 $ and at $ x=-4 $ , $ {f}’’(x)>0 $
Least value of $ f(x)=\frac{-4}{16+16}=-\frac{1}{8} $ .
BETA
