Functions — General Properties practice questions — Honors Math

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

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

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 $\mathbb{R}$.

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

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

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

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

(Note $g(f(x))=x^2-1$ — 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 $(f\circ g)(2)=9$.