The interval containing all the real values o | vedclass

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

Question

(A) Given function: $f(x) = \sqrt{x} + \frac{1}{\sqrt{x}}$.For $f(x)$ to be defined,we must have $x > 0$.Now,find the derivative $f'(x)$:$f'(x) = \fra

Vedclass · Accepted Answer

$(1, \infty)$

Vedclass · Answer

(A) Given function: $f(x) = \sqrt{x} + \frac{1}{\sqrt{x}}$.For $f(x)$ to be defined,we must have $x > 0$.Now,find the derivative $f'(x)$:$f'(x) = \frac{d}{dx} (x^{1/2} + x^{-1/2}) = \frac{1}{2} x^{-1/2} - \frac{1}{2} x^{-3/2}$.$f'(x) = \frac{1}{2\sqrt{x}} - \frac{1}{2x\sqrt{x}} = \frac{1}{2\sqrt{x}} (1 - \frac{1}{x}) = \frac{x-1}{2x\sqrt{x}}$.For the function to be strictly increasing,we require $f'(x) > 0$.Since $x > 0$,the denominator $2x\sqrt{x}$ is always positive.Therefore,$f'(x) > 0$ if and only if $x - 1 > 0$,which implies $x > 1$.Thus,the function is strictly increasing in the interval $(1, \infty)$.

The interval containing all the real values of $x$ such that the real-valued function $f(x) = \sqrt{x} + \frac{1}{\sqrt{x}}$ is strictly increasing is

Similar Questions

In which interval is the given function $f(x) = 2x^3 - 15x^2 + 36x + 1$ monotonically decreasing?

Find the intervals in which the function $f$ given by $f(x) = 4x^3 - 6x^2 - 72x + 30$ is
$(a)$ increasing
$(b)$ decreasing.

If $f(x) = \int_{x^2}^{x^2 + 1} e^{-t^2} dt$,then $f(x)$ increases in

Show that the function given by $f(x) = \sin x$ is neither increasing nor decreasing in $(0, \pi)$.

In the open interval $\left(0, \frac{\pi}{2}\right)$,which of the following is true for the expression $\cos x + x \sin x$?

Vedclass Test Series

Exam Paper Generator

Online Exam Module