I am trying to understand 1-1 and onto. I want to do this by attempting to prove that a function $f: \mathbb{R} \rightarrow \mathbb{R}$ defined by $x^3 +2$ is 1-1 and onto.

My attempt:

To prove that a function is $1$-$1$, we can show that if two elements in the domain can be mapped to the same element in the co-domain, then they are the same element.

The domain and co-domain for this example are the same ($\mathbb{R}$).

Suppose $x$ and $y$ are real numbers and suppose that $f(x)=f(y)$. We need to show that $x=y$. Since we said $f(x)=f(y)$:

$$x^3+2 = y^3+2$$ $$x^{1/3} = y^{1/3}$$ $$\pm x=\pm y$$ $$x=y$$ $$\Rightarrow \text{one-to-one}$$

To prove a function is onto, we need to show that for every real number $y$, there exists a real number $x$ such that $f(x)=y$. Let $x= (y-2)^{1/3}$, now $f(x)=(y-2)^{(1/3)(3)}+2=y-2+2=y \Rightarrow \text{onto}$.

My first question: Is my proof right?

My second question: Is there a way to just know (by intuition, or maybe graphically) that it's obvious that such function is one-to-one and onto?


Hint: (For second question)

A polynomial is bijective (one-one and onto) if and only if its derivative never changes sign.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.