Draw the graph of `f(x)=4x^(3)-3x` and hence draw the graph of `g(x)=cos^(-1)(4x^(3)-3x)`.

Explanation

`f(x)=4x^(3)-3x`
`:. f'(x)=12x^(2)-3`
`f'(x)=0 :. x=+-1//2`
Also `f(x)` is an odd function.
`x=1//2` is the point of minima and `x=-1//2` is the point of maxima.

Now for `g(x) = cos^(-1)(4x^(3)-3x)=cos^(-1)f(x)`,
`4x^(3)-3x in [-1, 1]` for which `x in [-1, 1]`
So we consider the following graph of `y=f(x)` for ` x in [-1, 1]`.

Now `g(-1)=cos^(-1)f(-1)=cos^(-1)(-1)=pi`
`g(-1//2)=cos^(-1)f(-1//2)=cos^(-1)(1)=0`
`g(0)=cos^(-1)f(0)=cos^(-1)(0)=pi//2`
`g(1//2)=cos^(-1)f(1//2)=cos^(-1)(-1)=pi`
`g(1)=cos^(-1)f(1)=cos^(-1)(1)=0`
Also `g'(x)=(f'(x))/(sqrt(1-(f(x))^(2)))`
So `g(x)` in non-differentiable at `x=+-1//2`.
Hence the graph of `g(x)=cos^(-1)(4x^(3)-3x)` is as shown in the following figure.