The function f(x)=displaystyle2/x+5,\ x≠ 0 is

The correct answer is Option (4) → decreasing function for \( x\in\mathrm{R}-\{0\} \)

Whether a function is increasing or decreasing depends on the sign of its derivative: f′(x) > 0 means increasing, f′(x) < 0 means decreasing.

    \( \displaystyle f(x)=\frac{2}{x}+5 \)

    \( \displaystyle f'(x)=-\frac{2}{x^2} \)

For every x ≠ 0, \( x^2 \) is positive, so

    \( f'(x)\lt 0\ \text{for all}\ x\in\mathrm{R}-\{0\} \)

So f is decreasing at every point where it is defined. The point x = 0 is not in the domain, so the domain has two parts, \( (-\infty,0) \) and \( (0,\infty) \), and f decreases on each part:

    \( x=1,\ 2,\ 4:\quad f=7,\ 6,\ 5.5 \)
    \( x=-4,\ -2,\ -1:\quad f=4.5,\ 4,\ 3 \)

Here “decreasing for \( x\in\mathrm{R}-\{0\} \)” means decreasing on each of the two parts. Across the break at x = 0 the values do not keep decreasing: for example, −1 < 1 but f(−1) = 3 is less than f(1) = 7.

Hence f is a decreasing function for \( x\in\mathrm{R}-\{0\} \), on each part of its domain.