Draw the graph of `y=-cot^(-1)x`.

Explanation

We have `f(x)=cot^(-1)((2-|x|)/(2+|x|))`
`=pi/2-tan^(-1)((2-|x|)/(2+|x|))`
`=pi/2-tan^(-1)((1-|x|/2)/(1-|x|/2))`
`=pi/2-(tan^(-1)1-tan^(-1),|x|/2)`
`=pi/4+tan^(-1),|x|/2`
Now f(x) is an even function.
We need to draw the graph for `xge0`
For `xge0,f(x)=pi/4+tan^(-1),x/2`
`Now f(0)=pi/4`
`f'(x)=1/(2(1+x^(2)/4))gt0" for all x "gt0`
Hence f(x) is an incereasing function.
`f'(x)=x/(4(1+x^(2)/4))lt0" for all "xlt0`
Hence, the graph is concave downwarod.
Also when `xtooo " "f(x)to pi/4+pi/2=(3pi)/4`
From the above discussion, the graph of y=f(x) for `xgt0` is as shown in the following figure

Since the function is even, the graph for all real values of x is as shown in the following figure.