Draw the graph of `y=(3x-x^(3))/(1-3x^(2))` and hence the graph of `y=tan^(-1).(3x-x^(3))/(1-3x^(2))`.

Explanation

We have `y=f(x)=tan^(-1)((3x-x^(3))/(1-3x^(2)))`
Domain of f(x) is `R={-1/sqrt3,1/sqrt3}`
Let x `= tan theta, theta in (-pi//2, pi//2)`
`rArr" "theta=tan^(-1)x`
Now `tan^(-1)((3x-x^(3))/(1-3x^(2)))=tan^(-1)((3tantheta-tan^(3)theta)/(1-3tan^(2)theta))`
`=tan^(-1)(tan3theta)`
`tan^(-1)(tanalpha),"where "alphain(-3pi//2,3pi//2)`
Now consider the graph of `y=tan^(-1)(tan alpha)," where "alpha in (-3pi//2,3pi//2)`.

From the graph,
`tan^(-1)((3x-x^(3))/(1-3x^(2)))=tan^(-1)(tan alpha)`
`={{:(alpha+pi,-3pi//2ltalpha-pi//2),(alpha,-pi//2ltalphaltpi//2),(alpha-pi,pi//2ltalphalt3pi//2):}`
`={{:(3 tan^(-1)x+pi,-3pi//2lttan^(-1)xlt-pi//2),(3 tan^(-1)x,-pi//2lt3tan^(-1)xltpi//2),(3tan^(-1)x-pi,pi//2lt3tan^(-1)xlt3pi//2):}`
`={{:(3 tan^(-1)x+pi,-pi//2lttan^(-1)xlt-pi//6),(3 tan^(-1)x,-pi//6lttan^(-1)xltpi//6),(3tan^(-1)x-pi,pi//6lttan^(-1)xltpi//2):}`
`={{:(3 tan^(-1)x+pi,-ooltxlt-1//2),(3 tan^(-1)x,-1//sqrt3ltxlt1//sqrt3),(3tan^(-1)x-pi,1//sqrt3ltxltoo):}`
`tan^(-1)` is an invreasing function for `x in R`.
So all brance functions in (i) are increasing functions
`underset(xto-oo)(lim)(3 tan^(-1)x+pi)=-pi/2,underset(xto-1/sqrt3)(lim)(3 tan^(-1)x+pi)=pi/2`
`3tan^(-1)x+pi=0:.x=-sqrt3`
Thus, `tan^(-1)((3x-x^(3))/(1-3x^(2)))" increases from "-pi/2"to"pi/2"when x increases from "-oo"to"pi/2"intersecting the x-axis at x ="-sqrt3`
`underset(xtooo)(lim)(3tan^(-1)x-pi)=pi/2,underset(xto1/sqrt3)(lim)(3tan^(-1)x-pi)=-pi/2`
`3 tan^(-1)x-pi=0:.x=-sqrt3`
Thus `tan^(-1)((3x-x^(2))/(1-3x^(2)))" increases form "-pi/2"to"pi/2" when x increases form "1/sqrt3" to " oo "intersecting the x-axis at x"=sqrt3`
From this information, we can draw the graph of `y=tan^(-1)((3x-x^(2))/(1-3x^(2)))` can be drawn as follows.