Functions — General Properties

$g(3)=3^2=9$.

Then $f(9)=2\cdot 9+1=19$.

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

$f(g(x))=f(x-1)=(x-1{)}^2$.

(Note $g(f(x))=x^2-1$ — composition order matters.)

$f(f(x))=f(2x)=2\cdot (2x)=4x$.

We need $x-2\ge 0$, i.e. $x\ge 2$.

The value at $x=2$ is $\sqrt{0}=0$, which is allowed, so the domain is $[2,\mathrm{\infty })$.

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

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

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

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 $\mathbb{R}\to \mathbb{R}$.

$2x+1=y⟺x={\displaystyle \frac{y-1}{2}}$.

Every real $y$ has exactly one preimage, so $f$ is both injective and surjective, i.e. bijective.

For any $y\ge 0$, $y=(\sqrt{y}{)}^2$, so $f$ is surjective onto $[0,\mathrm{\infty })$.

But $f(2)=f(-2)=4$, so $f$ is not injective. Hence surjective but not injective.