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The function $f(x)=\frac{x}{3}+\frac{3}{x}$ decreases in the interval
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2704 Upvotes
Verified Answer
The correct answer is:
$(-3,3)$
Given,
$f(x)=\frac{x}{3}+\frac{3}{x}$
On differentiating w.r.t. $x$, we get
$f^{\prime}(x)=\frac{1}{3}-\frac{3}{x^{2}}
$Since, function is decreasing
$\begin{array}{ll}
\text { i.e., } & f^{\prime}(x) < 0 \\
\Rightarrow & \frac{1}{3}-\frac{3}{x^{2}} < 0 \\
\Rightarrow & \frac{1}{3} < \frac{3}{x^{2}} \\
\Rightarrow & x^{2} < 9 \\
\Rightarrow & x \in(-3,3)
\end{array}$
which is the required interval in which function $f(x)$ decreases.
$f(x)=\frac{x}{3}+\frac{3}{x}$
On differentiating w.r.t. $x$, we get
$f^{\prime}(x)=\frac{1}{3}-\frac{3}{x^{2}}
$Since, function is decreasing
$\begin{array}{ll}
\text { i.e., } & f^{\prime}(x) < 0 \\
\Rightarrow & \frac{1}{3}-\frac{3}{x^{2}} < 0 \\
\Rightarrow & \frac{1}{3} < \frac{3}{x^{2}} \\
\Rightarrow & x^{2} < 9 \\
\Rightarrow & x \in(-3,3)
\end{array}$
which is the required interval in which function $f(x)$ decreases.
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