Let $f(x): \mathbb{R} \to \mathbb{R}$ be the map determined by:
$f(x)=\begin{cases} x &x \ge 2\\ \frac{x^3}{4} &-1\le x < 2\\ x &x < -1\end{cases}$
Is there an easy way to determine if this function is injective and surjective?
Injectivity means that $f(x) = f(y)$, hence for both $x>2$ and $x< -1$, $x = y$. Is the same true for $x^3\over4$?
As for surjectivity, I couldn't figure it out.
injectivity: If $x \geq 2$, then $f(x) = x \geq 2$. Moreover, if $f(x_1) = f(x_2)$, then $f(x_1) = x_1 = x_2 = f(x_2)$.
If $x < - 1$, then $f(x) = x < -1$. Moreover, if $f(x_1) = f(x_2)$, then $f(x_1) = x_1 = x_2 = f(x_2)$.
It remains to establish that if $-1 \leq x < 2$, then $-1 \leq x < 2$ and that, if this is the case, then $f(x_1) = f(x_2) \implies x_1 = x_2$.
We will show that $f$ is strictly increasing on this interval. Suppose $x_1, x_2 \in [-1, 2)$ and $x_1 > x_2$. Then \begin{align*} f(x_1) - f(x_2) & = \frac{x_1^3}{4} - \frac{x_2^3}{4}\\ & = \frac{1}{4}(x_1^3 - x_2^3)\\ & = \frac{1}{4}(x_1 - x_2)(x_1^2 + x_1x_2 + x_2^2)\\ & = \frac{1}{4}(x_1 - x_2)\left(x_1^2 + x_1x_2 + \frac{1}{4}x_2^2 + \frac{3}{4}x_2^2\right)\\ & = \frac{1}{4}(x_1 - x_2)\left[\left(x_1 + \frac{x_2}{2}\right)^2 + \frac{3}{4}x_2^2\right]\\ & > 0 \end{align*} Thus, $f$ is strictly increasing on $[-1, 2)$. Moreover, if $x \in [-1, 2)$, then $$f(-1) = \frac{(-1)^3}{4} = -\frac{1}{4} \leq x < \frac{2^3}{4} = 2$$ so, if $x \in [-1, 2)$, then $-1 < f(x) < 2$.
Consequently, $f$ is injective.
surjectivity: Since $f: \mathbb{R} \to \mathbb{R}$, for the function to be surjective, we require that every real number be in the range. We have shown that \begin{align*} x \geq 2 & \implies f(x) \geq 2\\ -1 \leq x < 2 & \implies -\frac{1}{4} \leq x < 2\\ x < -1 & \implies f(x) < -1 \end{align*} Thus, $f(x) \notin [-1, -1/4)$. Hence, $f$ is not surjective.
The results above can be obtained more easily by sketching the function's graph. Since a horizontal line crosses the graph at most once, the function is injective. Since the horizontal line $y = -1/2$ does not cross the graph, the function is not surjective.