Question

which of the following is an increasing function?
$f(x)=1-\sqrt{x}$
$f(x)=\frac{1}{\sqrt{x}}$
$f(x)=-\sqrt{x^{2}}$
$f(x)=1+\sqrt{x}$

Explanation:

Step1: Analyze \(f(x)=1 - \sqrt{x}\)

The derivative of \(y = 1-\sqrt{x}=1 - x^{\frac{1}{2}}\). Using the power rule \((x^n)^\prime=nx^{n - 1}\), we have \(f^\prime(x)=-\frac{1}{2\sqrt{x}}\lt0\) for \(x\gt0\). So it is a decreasing function.

Step2: Analyze \(f(x)=\frac{1}{\sqrt{x}}=x^{-\frac{1}{2}}\)

Using the power rule \((x^n)^\prime=nx^{n - 1}\), \(f^\prime(x)=-\frac{1}{2}x^{-\frac{3}{2}}=-\frac{1}{2x\sqrt{x}}\lt0\) for \(x\gt0\). So it is a decreasing function.

Step3: Analyze \(f(x)=-\sqrt{x^{2}}=-|x|\)

When \(x\gt0\), \(f(x)=-x\), \(f^\prime(x)=- 1\lt0\); when \(x\lt0\), \(f(x)=x\), \(f^\prime(x)=1\gt0\). But the domain of the original problem (assuming we are comparing with the other functions which are mainly for non - negative \(x\) in the context of \(\sqrt{x}\)) and overall it is not an increasing function on a single interval.

Step4: Analyze \(f(x)=1+\sqrt{x}=1 + x^{\frac{1}{2}}\)

Using the power rule \((x^n)^\prime=nx^{n - 1}\), \(f^\prime(x)=\frac{1}{2\sqrt{x}}\gt0\) for \(x\gt0\).

Answer:

\(f(x)=1+\sqrt{x}\)