Question

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:

• A. $(1, \infty)$
• B. $(0, 1)$
• C. $(-\infty, 0) \cup (1, \infty)$
• D. $(-\infty, 0)$

Hint:

To find the increasing or decreasing nature of a function, compute $f'(x)$ and check where it is positive or negative.

Answer 1

The Correct Option is A

Step 1: Compute the First Derivative
Differentiate $f(x)$: $$f'(x) = \frac{d}{dx} \left( \sqrt{x} + \frac{1}{\sqrt{x}} \right).$$ $$f'(x) = \frac{1}{2\sqrt{x}} - \frac{1}{2x^{3/2}}.$$ $$f'(x) = \frac{1}{2} \left( \frac{1}{\sqrt{x}} - \frac{1}{x^{3/2}} \right).$$ $$f'(x) = \frac{1}{2} \times \frac{x - 1}{x^{3/2}}.$$ Step 2: Find When $f'(x)>0$
For the function to be strictly increasing: $$\frac{(x - 1)}{x^{3/2}}>0.$$ Since $x^{3/2}$ is always positive for $x>0$, the sign of $f'(x)$ depends on $(x - 1)$.
Step 3: Find the Interval Where $f'(x)>0$
- $x - 1>0$ when $x>1$, so $f'(x)>0$.
- $x - 1<0$ when $0<x<1$, so $f'(x)<0$, meaning $f(x)$ is decreasing in this region.
Step 4: Conclusion
Thus, the function is strictly increasing for: $$\mathbf{(1, \infty)}.$$