f(x) = frac{x}{2} + frac{2}{x}, x neq 0 is st | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(A) Given function is $f(x) = \frac{x}{2} + \frac{2}{x}$.To find the intervals where the function is strictly decreasing,we find the derivative $f'(x)

Vedclass · Accepted Answer

$(-2, 0) \cup (0, 2)$

Vedclass · Answer

(A) Given function is $f(x) = \frac{x}{2} + \frac{2}{x}$.To find the intervals where the function is strictly decreasing,we find the derivative $f'(x)$.$f'(x) = \frac{d}{dx} (\frac{x}{2} + 2x^{-1}) = \frac{1}{2} - 2x^{-2} = \frac{1}{2} - \frac{2}{x^2}$.For the function to be strictly decreasing,we require $f'(x) $\frac{1}{2} - \frac{2}{x^2} Since $2x^2 > 0$ for all $x 
eq 0$,the inequality holds when $x^2 - 4 $(x - 2)(x + 2) This inequality holds for $x \in (-2, 2)$.Since $x 
eq 0$,the function is strictly decreasing in the interval $(-2, 0) \cup (0, 2)$.

$f(x) = \frac{x}{2} + \frac{2}{x}, x \neq 0$ is strictly decreasing in

Similar Questions

What type of function is $y = x^4$?

The function $f(x) = e^{-1/x}$ is strictly increasing for all $x$ where

For what values of $x$ does the function $f(x) = (x - 1)^2 (x - 2)$ monotonically decrease?

Let $f: R \rightarrow R$ be defined as $f(x) = \begin{cases} -\frac{4}{3}x^3 + 2x^2 + 3x, & x > 0 \\ 3xe^x, & x \leq 0 \end{cases}$. Then $f$ is an increasing function in the interval:

If $y = 2x + \cot^{-1} x + \log(\sqrt{1 + x^2} - x)$,then $y$

Vedclass Test Series

Exam Paper Generator

Online Exam Module