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: \[ \mathbf{(1, \infty)}. \]