The interval containing all the real values o | vedclass

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(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

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$(1, \infty)$

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(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

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