Question

The function \(f(x)=\frac{2}{x}+5,\ x\ne 0\) is

• A. increasing function for \(x\in\mathbb{R}-\{0\}\)
• B. increasing function for \(x\in(0,\infty)\)
• C. neither increasing nor decreasing function for \(x\in(0,\infty)\)
• D. decreasing function for \(x\in\mathbb{R}-\{0\}\)

Hint:

\(f'(x)=-2/x^2<0\) for all \(x\ne 0\).

Answer 1

The Correct Option is D

Step 1: Find the derivative.
Write \(f(x)=2x^{-1}+5\). Then
\[ f'(x)=-\frac{2}{x^2} \]

Step 2: Find its sign.
For every \(x\ne 0\), \(x^2>0\), so \(f'(x)=-\frac{2}{x^2}<0\). The derivative is negative at every point of the domain.

Step 3: Apply the rule.
A function with \(f'(x)<0\) at every point of its domain is a decreasing function there. So \(f\) is decreasing for \(x\in\mathbb{R}-\{0\}\), and in particular on \((0,\infty)\) and on \((-\infty,0)\).

Step 4: Check options 1 and 2.
These say increasing, but \(f'<0\), so both are wrong.

Step 5: Check option 3.
It says neither increasing nor decreasing on \((0,\infty)\). But \(f'<0\) there, so the function is decreasing. It is wrong.

Step 6: Choose option 4.
Option 4 states that \(f\) is decreasing on the whole domain, which matches \(f'(x)<0\) for all \(x\ne0\). This is the intended reading of the question.

Final Answer:
The function is decreasing for \(x\in\mathbb{R}-\{0\}\), option 4. \[ \boxed{\text{decreasing for } x\in\mathbb{R}-\{0\}} \]