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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.

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  • $\begingroup$ Hint: what happens as $x\to \pm \infty$? $\endgroup$
    – lulu
    Apr 18 at 13:09
  • $\begingroup$ 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) $\endgroup$
    – Henry
    Apr 18 at 13:19
  • $\begingroup$ 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 $(x-1)^3 + 2$. $\endgroup$
    – Salcio
    Apr 18 at 13:20
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$$ f(x) = (x-1)^3+2 $$ $$ f'(x) = 3(x-1)^2 \ge 0 ~~\forall ~~x \in \mathbb{R} $$ 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 $f(x)$

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  • $\begingroup$ So I cannot show with using f(root of cube(y-2)+1)=y? there is no way that I can show f(x)=y directly? $\endgroup$
    – Marinette
    Apr 18 at 16:29
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    $\begingroup$ @Marinette You can! The statement $f(\sqrt[3]{y-2} + 1) = y$ is true, and will show that $f$ is surjective as well. $\endgroup$ Apr 18 at 16:45
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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.

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