Question

Find the domain of the function `(x^3-3x+2)^(1/2)`?

Answer 1

The domain of `f(x)` is `x in [-2,+oo)`

Explanation:

Let

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

Therefore,

`x^3-3x+2>=0`

Let

`g(x)=x^3-3x+2`

`g(1)=1-3+2=0`

So `(x-1)` is a factor of `g(x)`

We perform a long division

`color(white)(aaaa)` `x-1` `color(white)(aaaa)` `|` `x^3+0x^2-3x+2` `color(white)(aaaa)` `|` `x^2+x-2`

`color(white)(aaaaaaaaaaaaaaa)` `x^3-x^2`

`color(white)(aaaaaaaaaaaaaaaa)` `0+x^2-3x`

`color(white)(aaaaaaaaaaaaaaaaaa)` `+x^2-x`

`color(white)(aaaaaaaaaaaaaaaaaaa)` `+0-2x+2`

`color(white)(aaaaaaaaaaaaaaaaaaaaaaa)` `-2x+2`

`color(white)(aaaaaaaaaaaaaaaaaaaaaaaaaa)` `0+0`

So,

`g(x)=(x-1)(x^2+x-2)=(x-1)(x-1)(x+2)=(x-1)^2(x+2)`

We can build the sign chart

`color(white)(aaaa)` `x` `color(white)(aaaaa)` `-oo` `color(white)(aaaaaa)` `-2` `color(white)(aaaaa)` `-1` `color(white)(aaaa)` `+oo`

`color(white)(aaaa)` `x+2` `color(white)(aaaaaa)` `-` `color(white)(aaaa)` `0` `color(white)(aaa)` `+` `color(white)(aaaa)` `+`

`color(white)(aaaa)` `(x-1)^2` `color(white)(aaaa)` `+` `color(white)(aaaaa)` `color(white)(aaa)` `+` `color(white)(aa)` `0` #color(white) (a)# `+`

`color(white)(aaaa)` `g(x)` `color(white)(aaaaaaaa)` `-` `color(white)(aaa)` `0` `color(white)(aaa)` `+` `color(white)(aa)` `0` #color(white) (a)# `+`

Therefore,

`g(x)>=0` when `x in [-2,+oo)`

The domain of `f(x)` is `x in [-2,+oo)`

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