# How to show that $f(x)=(x − 1)^3 + 2$ is surjective?

I need to show that the function is bijective. I already showed that it is injective but I struggle to show that it is surjective

A function $$F\colon X\to Y$$ is called surjective, if for all $$y$$ belongs to $$Y$$, there exists an $$x$$ belongs to $$X$$ such that $$f(x)=y$$.

\begin{align}&y=(x-1)^3+2\\ &y-2=(x-1)^3\\ &\sqrt[3]{y-2}=x-1\\ &\sqrt[3]{y-2}+1=x\end{align}

I know that after, I need to set what I have in $$f(x)$$ but I have some problems to finish it.

• Hint: what happens as $x\to \pm \infty$?
– lulu
Apr 18 at 13:09
• You need to check that $\sqrt[3]{y-2}+1$ is well defined for all $y$: note that on the real numbers $\sqrt[2]{y-2}+1$ would not be for $y< 2$. Then show $f(\sqrt[3]{y-2}+1)=y$ for all $y$ (essentially go up your equations) Apr 18 at 13:19
• Look from geometric side rather than algebraic - $x^3$ is surjective, so if you move its graph by $1$ along x-axis, you still have surjective function, that is $(x-1)^3$. Then you move by 2 units along y-axis, still surjective, but it is $(x-1)^3 + 2$. Apr 18 at 13:20

$$f(x) = (x-1)^3+2$$ $$f'(x) = 3(x-1)^2 \ge 0 ~~\forall ~~x \in \mathbb{R}$$ Thus, $$f(x)$$ is an increasing function and is obviously continuous. Also, $$\lim_{x \to -\infty}f(x) = -\infty$$ $$\lim_{x \to \infty}f(x) = \infty$$ Now, you can safely conclude that all real numbers lie in the range of $$f(x)$$
• @Marinette You can! The statement $f(\sqrt[3]{y-2} + 1) = y$ is true, and will show that $f$ is surjective as well. Apr 18 at 16:45
Hint: Write $$f=g \circ h \circ j$$, where $$j(x)=x-1$$, $$h(x)=x^3$$, $$g(x)=x+2$$. Prove that $$g,h,j$$ are bijections.