1/x

X≠0 is available, so the zero point of g(x) is exactly the same as the non-zero point of xg(x).

So we consider the zero point of xg(x) =xf(x)+ 1.

Because when x≠0, f' (x)+f (x)/x.

>0,

① when x > 0, (XG (x))' = (xf (x))' = xf' (x)+f (x) = x (f' (x)+

f(x) /x )>0，

Therefore, at (0, +∞), the function xg(x) monotonically increases.

It's also VIII

Latent Image Memory (abbreviation for latent image memory)

x→0

[xf(x)+ 1]= 1, ∴ On (0, +∞), function XG (x) = xf (x)+ 1 > 1 holds.

So on (0, +∞), the function xg(x)=xf(x)+ 1 has no zero.

② when x < 0, because (XG (x))' = (xf (x))' = xf' (x)+f (x) = x (f' (x)+

F(x) /x) 1 hold,

Therefore, the function xg(x) has no zero at (-∞, 0).

To sum up, the function g (x) = f (x)+1/x.

The number of zeros on r is 0,

So choose C.