Are the following formula 1-1 or Onto Functions?

Question

1) Is the function Cube Root of $\sqrt[3]{{-6x-4}}$ One to One Function if domain is all real number?

IMO, I am assuming this is an 1-1 function because well, 1) This will produce a graph of square root. So every x will have a different y value.

That's my assumption, I am not too sure if my reasoning is correct. I know i would have to solve this function to find out if its really a 1-1 function but I don't know how to do it. My best try is as follows:

$F(x_1) = \sqrt[3]{{-6x-4}}$

$F(x_2) = \sqrt[3]{{-6x-4}}$

Then I would have $F(x_1) = F(x_2)$. I don't know what to do after this. It would be great if someone can tell me how to move forward with or how to solve this.

B) Is the Function $g(x) = {3x^x-9}$ onto function if the domain is also all Real Numbers?

Thank you.

P.S. i don't how to write it properly. So can someone edit it please.

Answer 1

We consider $f:\Bbb R\to\Bbb R$ defined by $f(x)=\sqrt[3]{-6x-4}$. Since $\sqrt[3]{-1}=-1$ (over the reals), we have $f(x)=-\sqrt[3]{6x+4}$. To show that $f$ is one-to-one, we need to show that $$f(x)=f(y)\implies x=y$$ So we need to show that

$$\sqrt[3]{6x+4}=\sqrt[3]{6y+4}\implies x=y$$

To show that $f$ is onto we have to show that, for any $y\in\Bbb R$, there exists $x\in\Bbb R$ such that $f(x)=y$. So we need to show that

$$\text{For any $y\in\Bbb R$, there exists $x\in\Bbb R$ such that }~ -\sqrt[3]{6x+4}=y$$