For what values of x is the function f(x) = x | vedclass

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

Question

(B) Given the function $f(x) = x + \frac{1}{x}$.To find where the function is strictly increasing,we calculate the derivative $f'(x)$ and set $f'(x) >

Vedclass · Accepted Answer

$|x| > 1$

Vedclass · Answer

(B) Given the function $f(x) = x + \frac{1}{x}$.To find where the function is strictly increasing,we calculate the derivative $f'(x)$ and set $f'(x) > 0$.$f'(x) = \frac{d}{dx}(x + x^{-1}) = 1 - x^{-2} = 1 - \frac{1}{x^2}$.For strictly increasing,$f'(x) > 0$,so $1 - \frac{1}{x^2} > 0$.$\frac{x^2 - 1}{x^2} > 0$.Since $x^2 > 0$ for all $x 
eq 0$,the inequality holds if $x^2 - 1 > 0$.$x^2 > 1$,which implies $|x| > 1$.

For what values of $x$ is the function $f(x) = x + \frac{1}{x}, x \neq 0$ strictly increasing?

Similar Questions

If $f(x) = x^5 - 20x^3 + 240x$,then $f(x)$ satisfies which of the following?

Consider the following statements $S$ and $R$:
$S$: Both $\sin x$ and $\cos x$ are decreasing functions in $\left( \frac{\pi}{2}, \pi \right)$.
$R$: If a differentiable function decreases in $(a, b)$,then its derivative also decreases in $(a, b)$.
Which of the following is true?

Show that the function given by $f(x) = 7x - 3$ is strictly increasing on $R$.

$y = x^2 (x - 3)^2$ increases for all values of $x$ lying in the interval :

For every value of $x \in [1, 3]$,the function $f(x) = \frac{1}{8^x}$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module