Q.Show that the function $f : \mathbb{R} \to \mathbb{R}$ defined by $f(x) = x^3$ is both one-one and onto.

One-one: suppose $f(x_1)=f(x_2)$, i.e. $x_1^3=x_2^3$. Since the cube function is strictly increasing over all of $\mathbb{R}$ (it never repeats a value, unlike $x^2$), this forces $x_1=x_2$. So $f$ is one-one.

Onto: for any $y\in\mathbb{R}$, the real cube root $x=y^{1/3}$ exists and is unique (unlike square roots, cube roots are defined for negative reals too), and $f(y^{1/3})=(y^{1/3})^3=y$. So every $y$ has a preimage, and $f$ is onto.

Since $f$ is both one-one and onto, it is bijective.

Use the strictly-increasing shortcut for injectivity, and the existence of a unique real cube root for every real number (positive, negative or zero) to establish surjectivity.

• Confusing $x^3$ with $x^2$ and wrongly concluding it is not one-one.
• Forgetting that, unlike square roots, cube roots of negative numbers are real, so onto genuinely holds here.
• Treating 'odd power' functions all the same without checking the specific function stated.