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Please solve RD Sharma class 12 chapter 16 Increasing and Decreasing Function Excercise Fill in the blanks Question 3: Maths Textbook Solution.

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Answer: The function g(x) decreases in the closed intervals [-1,1]

Hint: If f(x) decreasing, f'(x)<0

Given:  g(x)=x+\frac{1}{x}, x\neq 0

Explanation: It is given that



So, 1-\frac{1}{x^{2}}<0                   [  \because g(x) decreasing]

\Rightarrow \frac{x^{2}-1}{x^{2}}<0

\Rightarrow \frac{(x-1)(x+1)}{x^{2}}<0

\because x\neq 0\Rightarrow (x-1)(x+1)<0

\Rightarrow x\epsilon [-1,1]

Thus, the function decreases in [-1,1]

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