Applications Of Derivatives Question 284
Function $ f(x)=\frac{4x^{2}+1}{x} $ is decreasing on interval
Options:
A) $ ( \frac{-1}{2},,\frac{1}{2} ) $
B) $ [ \frac{1}{2},,-\frac{1}{2} ] $
C) (- 1, 1)
D) [1, 1]
Show Answer
Answer:
Correct Answer: B
Solution:
$ f(x)=4x+\frac{1}{x} $
$ \frac{d}{dx}f(x)=\frac{d}{dx}[ 4x+\frac{1}{x} ] $ = $ 4-\frac{1}{x^{2}} $
For extremum, $ \frac{dy}{dx}=0 $
therefore $ 4-\frac{1}{x^{2}}=0 $
therefore $ x=\frac{1}{2},,-\frac{1}{2} $
$ f\ ( \frac{1}{2} )=4.\frac{1}{2}+\frac{1}{1/2} $ = $ 2+4=6 $
$ f\ ( -\frac{1}{2} )=4.( -\frac{1}{2} )+\frac{1}{-1/2}=-2-2=-4 $
Hence $ f(x) $ is decreasing in interval $ [-1/2,,1/2] $ .
BETA
