Draw the graph of `y=f(x)=(x+1)/(x^(2)+1)`

Explanation

The domain of the function is R as `x^(2)+1 !=0` for any real number.
1. y-intercept
`f(0)=1`.
2. x-intercept (zeros)
Put `y=0 " or " x+1=0`
Hence the graph intersects x-axis at (-1, 0).
3 Asymptotes
Vertical asymptotes
Graph does not have vertical asymptote as `x^(2)+1 !=0` for any real number.
Horizontal asymptotes
Since the degree of the denominator is higher than the degree of denominator, the graph has horizontal asymptote y = 0.
Clearly, the graph has no oblique asymptote.
4. Extremum
`f'(x)=(-x^(2)-2x+1)/((x^(2)+1)^(2))`
`f'(x)=0 rArr x^(2)+2x-1=0` or `xx=-1+-sqrt(2)`
Sign scheme of `f'(x)` is as follows.

From the sign scheme, `x=-1-sqrt(2)` is the point of minima and `-1+sqrt(2)` is the point of maxima.
Also `f(-1-sqrt(2))=(-1-sqrt(2)+1)/((-1-sqrt(2))^(2)+1)=(-sqrt(2)+1)/(2)`
and `f(-1+sqrt(2))=(-1+sqrt(2)+1)/((-1+sqrt(2))^(2)+1)=(sqrt(2)+1)/(2)`
From the above discussion , we have the following reference points and lines.

When `xrarr-oo,yrarr0`, and the function decreases in `(-oo,0)`.
The function attains the minimum value at `x=-1-sqrt(2)` and then increases in `(-1-sqrt(2),-1+sqrt(2))` while intersecting the x-axis at (-1,0) and the y-axis at (0,1). It attains the maximum value at `x=-1+sqrt(2)` and then decreases.
When `xrarroo, y rarr 0`.
Thus, the graph of `y=f(x)` is as shown in the following figure.