Functions — General Properties practice questions — GCSE Maths

$f$ is strictly increasing on $\mathbb{R}$, hence injective. For every $y \in \mathbb{R}$, $x = \sqrt[3]{y}$ satisfies $f(x) = y$, hence surjective. So $f$ is bijective, with $f^{-1}(y) = \sqrt[3]{y}$.

Left branch ($x < 0$): $f(x) = x + 1 \in (-\infty,\, 1)$, strictly increasing. Right branch ($x \ge 0$): $f(x) = 2x + 1 \in [1,\, +\infty)$, strictly increasing.

The two images $(-\infty, 1)$ and $[1, +\infty)$ are disjoint and together cover all of $\mathbb{R}$, so $f$ is surjective. Each branch is strictly increasing, and the ranges don't overlap, so no two distinct $x$-values give the same $f(x)$ — $f$ is injective. Hence $f$ is bijective.

(Note: $f(-1) = 0$ and $f(0) = 1$, so distractor A is factually false.)

The graph of $f^{-1}$ is the reflection of the graph of $f$ across $y = x$. Their intersection lies on $y = x$, so we solve $f(x) = x$: $2x - 3 = x \Rightarrow x = 3$.

$f$ is invertible iff its slope $a - 2 \ne 0$. So $f$ fails to be invertible exactly when $a = 2$ (it then becomes the constant $5$, neither injective nor surjective).

$y = 3x - 6 \Rightarrow x = \dfrac{y + 6}{3}$, so $f^{-1}(y) = \dfrac{y + 6}{3}$.

$f(-1) = 1 = f(1)$, so $f$ is not injective. The value $-1$ has no real preimage (since $x^2 \ge 0$), so $f$ is not surjective either.

We need $x - 2 \ge 0$, i.e. $x \ge 2$. The domain is $[2, +\infty)$.

$f(-x) = (-x)^3 = -x^3 = -f(x)$.

Since $f(-x) = -f(x)$ for all $x$, the function is odd.

$f(1) = 1$, $f(2) = 3$, $f(3) = 5$.

The image is the set of outputs: $\{1, 3, 5\}$.

$g\big(f(x)\big) = g(x + 3) = 2(x + 3) = 2x + 6$.