KCET · Maths · Application of Derivatives
The function \(f(x)=\frac{x}{3}+\frac{3}{x}\) decreases in the interval
- A \((-3,3)\)
- B \((-\infty, 3)\)
- C \((3, \infty)\)
- D \((-9,9)\)
Answer & Solution
Correct Answer
(A) \((-3,3)\)
Step-by-step Solution
Detailed explanation
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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