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The interval in which the function $f(x)=$ $\frac{4 x^2+1}{x}$ is decreasing is :
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The correct answer is:
$\left(-\frac{1}{2}, \frac{1}{2}\right)$
Given $f(x)=\frac{4 x^2+1}{x}$ Thus $f^{\prime}(x)=4-\frac{1}{x^2}$
$\mathrm{f}(\mathrm{x})$ will be decreasing if $\mathrm{f}^{\prime}(\mathrm{x}) < 0$
Thus $4-\frac{1}{x^2} < 0 \Rightarrow \frac{1}{x^2}>4 \Rightarrow \frac{-1}{2} < x < \frac{1}{2}$
Thus interval in which $f(x)$ is decreasing, is $\left(-\frac{1}{2}, \frac{1}{2}\right)$.
$\mathrm{f}(\mathrm{x})$ will be decreasing if $\mathrm{f}^{\prime}(\mathrm{x}) < 0$
Thus $4-\frac{1}{x^2} < 0 \Rightarrow \frac{1}{x^2}>4 \Rightarrow \frac{-1}{2} < x < \frac{1}{2}$
Thus interval in which $f(x)$ is decreasing, is $\left(-\frac{1}{2}, \frac{1}{2}\right)$.
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