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

The two images $(-\mathrm{\infty },1)$ and $[1,+\mathrm{\infty })$ 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 $a-2\mathrm{\ne }0$. So $f$ fails to be invertible exactly when $a=2$ (it then becomes the constant $5$, neither injective nor surjective).

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

$f(-1)=1=f(1)$, 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 $[2,+\mathrm{\infty })$.

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

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: $\{1,3,5\}$.

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