Find the intervals in which f(x)=4 x^2+1 x is increasing or decre… - Vidyadip

Question

Find the intervals in which $f(x)=\frac{4 x^2+1}{x}$ is increasing or decreasing.

✓

Answer

Given: $f ( x )=\frac{4 x^2+1}{x}$
$\Rightarrow f(x)=4 x+\frac{1}{x}$
$\Rightarrow f^{\prime}(x)=4-\frac{1}{x^2}=\frac{4 x^2-1}{x^2}$
For $f(x)$ to be increasing, we must have
$f^{\prime}(x) > 0$
$\Rightarrow \frac{4 x^2-1}{x^2} > 0$
$\Rightarrow 4 x^2-1 > 0$
$\Rightarrow \quad(2 x-1)(2 x+1) > 0$
$(x-1 / 2)(x+1 / 2) > 0$
$x<-1 / 2 \text { or }, x >1 / 2$
$x \in(-\infty,-1 / 2) \cup(1 / 2, \infty)$
So , $f(x)$ is increasing on $(-\infty,-1 / 2) \cup(1 / 2, \infty)$
for $f(x)$ is to be decreasing, we must have
$\Rightarrow \frac{4 x^2-1}{x^2}<0$
$\Rightarrow 4 x^2-1 < 0$
$(2 x-1)(2 x+1) < 0$
$-\frac{1}{2}< x <\frac{1}{2}$
$x \in(-1 / 2,1 / 2)$
But domain $f R -\{0\}$.
So, $f(x)(-1 / 2,0) \cup(0,1 / 2)$

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