Question

f(x)=`x-3(x)^(2/3)+3(x)^(1/3)`,solve in R+ the equation f(x)=x ?

Answer 1

`x=1`

Explanation:

In `R^+`
`color(white)("XXX")x > 0`
(which implies `x!=0`)

If `f(x)=x-3(x)^(2/3)+3(x)^(1/3)`
and `f(x)=x`
then
`color(white)("XXX")x-3(x)^(2/3)+3(x)(1/3)=x`

Subtracting `x` from both sides:
`color(white)("XXX")-3(x)^(2/3)+3(x)^(1/3)=0`

Dividing both sides by `3` and moving the negative term to the right side:
`color(white)("XXX")x^(1/3)=x^(2/3)`
`color(white)("XXX"x^(1/3))=x^(1/3) *x^(1/3)`

Dividing both sides by `x^(1/3)` (we can do this because we know `x!=0`) and reversing the sides:
`color(white)("XXX")x^(1/3)=1`

Cubing both sides:
`color(white)("XXX")x=1`