Solveeit Logo

Question

Mathematics Question on Rate of Change of Quantities

The interval in which the function f(x)=4x2+1xf\left(x\right) = \frac{4x^{2}+1}{x} is decreasing is :

A

(12,12)\left(-\frac{1}{2}, \frac{1}{2}\right)

B

[12,12]\left[-\frac{1}{2}, \frac{1}{2}\right]

C

(1,1)(-1,1)

D

[1,1][-1,1]

Answer

(12,12)\left(-\frac{1}{2}, \frac{1}{2}\right)

Explanation

Solution

Given f(x)=4x2+1xf\left(x\right) = \frac{4x^{2}+1}{x} Thus f(x)=41x2f'\left(x\right) = 4-\frac{1}{x^{2}}
f(x)f\left(x\right) will be decreasing if f(x)<0f'\left(x\right) < 0
Thus 41x2<04-\frac{1}{x^{2}} < 0
1x2>412<x1<12\Rightarrow \frac{1}{x^{2}} > 4 \Rightarrow \frac{-1}{2} < x 1 < \frac{1}{2}
Thus interval in which f(x)f\left(x\right) is decreasing, is (12,12)\left(-\frac{1}{2}, \frac{1}{2}\right).