Functions — General Properties

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

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

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

$f\big(g(x)\big) = f(x - 1) = (x - 1)^2$.

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

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

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 $[2, \infty)$.

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

$x^2$, $|x|$ 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 $\mathbb{R} \to \mathbb{R}$.

$2x + 1 = y \iff x = \dfrac{y - 1}{2}$.

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 $[0, \infty)$.

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