Functions, domain, range and composition practice questions — IB Maths AA

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 $f(x) = x$: $2x - 3 = x \Rightarrow x = 3$.

$y = 3x - 6 \Rightarrow x = \dfrac{y + 6}{3}$, 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.

$f$ 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).

The slope is $-3 < 0$, so larger $x$ gives smaller $f(x)$.

Hence $f$ is strictly decreasing on $\mathbb{R}$.

The denominator vanishes when $x - 3 = 0$, i.e. $x = 3$.

Everywhere else the value exists, so the domain is $\mathbb{R} \setminus \{3\}$.

$f(2) = f(-2) = 4$, so $f$ is not injective.

No real $x$ gives $x^2 = -1$, so negative values are never reached: $f$ is not surjective onto $\mathbb{R}$. Hence neither.

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

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

$x^2$, $|x|$ and the constant $3$ each give one value for every real $x$.

$\pm\sqrt{x}$ fails twice: it is not single-valued (two outputs) and is undefined for $x < 0$. So it is not a function $\mathbb{R} \to \mathbb{R}$.

$f(-x) = (-x)^2 + 1 = x^2 + 1 = f(x)$.

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