Question

Analyse the function `f(x) = (1+x)^(2/3)*(3-x)^(1/3)` for extreme values, zeros and points of discontinuity, if any?

Answer 1

`f(x)` has a local maximum of `~=2.117` at `x=5/3`
`f(x)` has zeros at `x=-1` and `x=3`
`f(x)` has a discontinuity at `(-1,0)`

Explanation:

`f(x) = (1+x)^(2/3)*(3-x)^(1/3)`

By inspection, `f(x)` has zeros at `x=-1` and `x=3`

Applying the product rule:

`f'(x)` = `(1+x)^(2/3) * 1/3(3-x)^(-2/3) * (-1)+`
`2/3(1+x)^(-1/3) * (3-x)^(1/3)`

For points of extrema `f'(x)=0`

I.e. where: `(1+x)^(2/3) * 1/3(3-x)^(-2/3) * (-1)+`
`2/3(1+x)^(-1/3) * (3-x)^(1/3) =0`

`1/3(1=x)^(2/3) * (3-x)^(-2/3) = 2/3(1+x)^(-1/3) * (3-x)^(1/3)`

`(1+x)^(2/3)/((3-x)^(2/3)) = (2*(3-x)^(1/3))/(1+x)^(1/3)`

Cross multiply:

`(1+x)^(2/3+1/3) = 2* (3-x)^(1/3+2/3)`

`1+x = 2(3-x) -> 1+x = 6-2x`

`3x=5`

`x=5/3`

Hence `f(x)` has an extreme value at `x=5/3`

Now consider the graph of `f(x)` below:

graph{(1+x)^(2/3)* (3-x)^(1/3) [-10, 10, -5, 5]}

We can see that `f(x)` has a local maximum at `x=5/3`

`:. f_max(x) = f(5/3) ~= 2.117`

Finally, considering the zeros of `f(x)` we can see that `f(x)` has a discontinuity at `(-1,0)`