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

Explanation

`y=f(x)=tan^(-1)((1-x^2)/(1+x^(2)))`
`tan^(-1)1-tan^(-1)x^(2)`
`=pi/4-tan^(-1)x^(2)`
Clearly, the domain of the function is R and `X^(2)ge0,AA x in R.`
Also function is even therefore the graph is symmetrical about the y-axis.
`tan^(-1)x^(2)ge0thereforetan^(-1) x^(2) in [0, pi/2)`
`therefore" "tan^(-1)x^(2) in (pi/2,0]`
`therefore" "pi/4-tan^(-1)x^(2) in(-pi/4,pi/4]`
Lets us first draw the graph for `xge0.`
`f(0)=pi/4-tan^(-1)0=pi/4`
Now if we increase the value of x from x=0, we find that the value of `x^(2)` increases, so the value of `tan^(-1)x^(2)` increases. Hence the value of `pi/4-tan^(-1)x^(2)` decreases.
When `x to, pi/4-tan^(-1)x^(2)topi/4-pi/2=pi/4`
Also when `pi/4-tan^(-1)x^(2)=0, tan^(-1)x^(2)=pi/4thereforex=1`
Thus, the graph intersects the x-axis at (1,0).
Thus, the graph of `f(x)" for x"le0` is as whown in the following figure.

Since f(x) is an even function, the graph is symmetrical about the y-xis, heance the graph of the function is as shown in the following figure.

Here `f'(x) = 0 `0-0(2x)/(1+x^(2))`, therefore f(x) is differentiable at x = 0.
Also f"(x) `=2(x^(2)-1)/((1+x^(2))^(2))`
`f"(x) = 0 therefore x = +-1,` where the graph of f(x) changes its concavity.
For `x in (-1, 1), f"(x) lt0,` where tghe graph is concave downwards.
For `x in (-oo, -1)cup (1, oo), f"(x) lt0`, where the graph is convace upwards.