Question

which of the following is a decreasing function?

$f(x)=-sqrt{2 - x}$

$2-sqrt{x}$

$sqrt3{x}$

$f(x)=sqrt{x - 2}$

Explanation:

Step1: Analyze \(y =-\sqrt{2 - x}\)

Let \(u = 2 - x\), then \(y=-\sqrt{u}\). The domain of \(y =-\sqrt{2 - x}\) is \(x\leqslant2\). The function \(u = 2 - x\) is decreasing (\(u^\prime=- 1<0\)), and \(y =-\sqrt{u}\) (\(y^\prime=-\frac{1}{2\sqrt{u}}<0\) for \(u>0\)). By the chain - rule \(y^\prime=\frac{1}{2\sqrt{2 - x}}>0\) for \(x < 2\), so it is increasing.

Step2: Analyze \(y = 2-\sqrt{x}\)

The domain of \(y = 2-\sqrt{x}\) is \(x\geqslant0\). The derivative \(y^\prime=-\frac{1}{2\sqrt{x}}<0\) for \(x>0\).

Step3: Analyze \(y=\sqrt[3]{x}\)

The derivative \(y^\prime=\frac{1}{3x^{\frac{2}{3}}}>0\) for \(x
eq0\), and the function is increasing for all \(x\in R\).

Step4: Analyze \(y=\sqrt{x - 2}\)

The domain of \(y=\sqrt{x - 2}\) is \(x\geqslant2\). The derivative \(y^\prime=\frac{1}{2\sqrt{x - 2}}>0\) for \(x>2\), so it is increasing.

Answer:

\(2-\sqrt{x}\)