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

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

$f$ 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).

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

$x^2$, $\mathrm{\mid }x\mathrm{\mid }$ 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.