Functions — General Properties practice questions — Honors Math

$x^2$, $|x|$ 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.

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

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

The slope is $a = 2 > 0$, so as $x$ grows $f(x)$ grows.

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

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

$x^2$ and $|x|$ both give $f(1) = f(-1)$; $\sin x$ repeats every $2\pi$ — none injective.

$x^3$ is strictly increasing on $\mathbb{R}$, so different inputs give different outputs: it is injective.

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

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

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

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

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

$g(2) = 5$, then $f(5) = 2 \cdot 5 - 1 = 9$. Hence $(f \circ g)(2) = 9$.