Draw and discuss the graph of `f(x) = x^(2//3)-x^(4//3)`

Explanation

We have `f(x) = x^(2//3)-x^(4//3)`
Clearly, the domain of the function is R.
`f(x) = x^(2//3)(1-x^(2//3))=x^(2//3)(1-x^(1//3))(1+x^(1//3))`
Sign of scheme of `y=f(x)` is as follows:

`f^(')(x) = 2/3x^(-1/3) - 4/3x^(1/3)`
`=2/3(1-2x^(2//3))/(x^(1//3)`
`=-4/3(x^(1//3)-1/sqrt(2))(x^(1/3)+1/sqrt(2))/(x^(1/3))`
`f^(')(x)=0 therefore x =+-1/(2sqrt(2))`
Also, `f^(')(x)` does not exist at x=0
it is also a critical point.
Sign scheme of `f^(')(x)` is as follows:

Thus, `x=0` is the point of minima and `x = +-1/(2sqrt(2))` are the points of maxima.
`f(0)=0, f(x)=0 therefore x=0, +-1`
`f(+-1)=0`
Also when, `x to +-infty, f(x) to -infty`
`y=f(x)` is an even function, so the graph is symmetrical about the y-axis.
At `x=0`, the graph has vertical tangent line `x=0`, where the function is non-differentiable.
From the above discussion, the graph of the function is as shown in the following figure.