Operations on functions, composite and inverse functions

Let $f(x)=\sqrt{x-1}f(x)\; =\; \backslash sqrt\{x\; -\; 1\}$ and $g(x)=x^2-4g(x)\; =\; x^2\; -\; 4$. Give $f+gf\; +\; g$ and $\frac{f}{g}\backslash frac\{f\}\{g\}$ with their domains.

Solution $\mathrm{Dom}f=[1,+\mathrm{\infty })\backslash operatorname\{Dom\}\; f\; =\; [1,\; +\backslash infty)$, $\mathrm{Dom}g=\mathbb{R}\backslash operatorname\{Dom\}\; g\; =\; \backslash mathbb\{R\}$.

• $(f+g)(x)=\sqrt{x-1}+x^2-4(f\; +\; g)(x)\; =\; \backslash sqrt\{x\; -\; 1\}\; +\; x^2\; -\; 4$, domain $[1,+\mathrm{\infty })[1,\; +\backslash infty)$.
• $\left(\frac{f}{g}\right)(x)=\frac{\sqrt{x-1}}{x^2-4}\backslash left(\backslash frac\{f\}\{g\}\backslash right)(x)\; =\; \backslash frac\{\backslash sqrt\{x\; -\; 1\}\}\{x^2\; -\; 4\}$; $g(x)=0g(x)\; =\; 0$ at $x=\pm 2x\; =\; \backslash pm\; 2$, and only $22$ lies in $[1,+\mathrm{\infty })[1,\; +\backslash infty)$. Domain $[1,2)\cup (2,+\mathrm{\infty })[1,\; 2)\; \backslash cup\; (2,\; +\backslash infty)$.

(a) For $f(x)=2x+3f(x)\; =\; 2x\; +\; 3$ and $g(x)=x^{2}g(x)\; =\; x^2$, find $f\circ gf\; \backslash circ\; g$ and $g\circ fg\; \backslash circ\; f$.
(b) For $u(x)=\sqrt{x}u(x)\; =\; \backslash sqrt\{x\}$ and $v(x)=1-x^{2}v(x)\; =\; 1\; -\; x^2$, find $u\circ vu\; \backslash circ\; v$ and $v\circ uv\; \backslash circ\; u$ with their domains.

Solution

1. $(f\circ g)(x)=f(x^2)=2x^2+3(f\; \backslash circ\; g)(x)\; =\; f(x^2)\; =\; 2x^2\; +\; 3$; $(g\circ f)(x)=(2x+3{)}^2=4x^2+12x+9(g\; \backslash circ\; f)(x)\; =\; (2x\; +\; 3)^2\; =\; 4x^2\; +\; 12x\; +\; 9$. Different functions: composition is not commutative.
2. $(u\circ v)(x)=\sqrt{1-x^2}(u\; \backslash circ\; v)(x)\; =\; \backslash sqrt\{1\; -\; x^2\}$, which needs $1-x^2\ge 01\; -\; x^2\; \backslash ge\; 0$: domain $[-1,1][-1,\; 1]$.
   $(v\circ u)(x)=1-(\sqrt{x}{)}^2=1-x(v\; \backslash circ\; u)(x)\; =\; 1\; -\; (\backslash sqrt\{x\})^2\; =\; 1\; -\; x$, but only for $x\in \mathrm{Dom}u=[0,+\mathrm{\infty })x\; \backslash in\; \backslash operatorname\{Dom\}\; u\; =\; [0,\; +\backslash infty)$. The simplified formula $1-x1\; -\; x$ hides this restriction, so state it: $v\circ u(x)=1-xv\; \backslash circ\; u(x)\; =\; 1\; -\; x$, $x\ge 0x\; \backslash ge\; 0$.

Find the inverse of $f(x)=\frac{2x+1}{x-1}f(x)\; =\; \backslash frac\{2x\; +\; 1\}\{x\; -\; 1\}$, $x\mathrm{\ne }1x\; \backslash ne\; 1$.

Solution From Lesson 2, $y=\frac{2x+1}{x-1}⟺x=\frac{y+1}{y-2}y\; =\; \backslash frac\{2x\; +\; 1\}\{x\; -\; 1\}\; \backslash iff\; x\; =\; \backslash frac\{y\; +\; 1\}\{y\; -\; 2\}$ ($y\mathrm{\ne }2y\; \backslash ne\; 2$), and each $y\mathrm{\ne }2y\; \backslash ne\; 2$ comes from exactly one $xx$, so $ff$ is one-to-one onto $\mathbb{R}\setminus \{2\}\backslash mathbb\{R\}\; \backslash setminus\; \backslash \{2\backslash \}$. Hence

Check: $f(f^{-1}(x))=\frac{2\frac{x+1}{x-2}+1}{\frac{x+1}{x-2}-1}=\frac{\frac{3x}{x-2}}{\frac{3}{x-2}}=xf(f^\{-1\}(x))\; =\; \backslash frac\{2\backslash frac\{x\; +\; 1\}\{x\; -\; 2\}\; +\; 1\}\{\backslash frac\{x\; +\; 1\}\{x\; -\; 2\}\; -\; 1\}\; =\; \backslash frac\{\backslash frac\{3x\}\{x\; -\; 2\}\}\{\backslash frac\{3\}\{x\; -\; 2\}\}\; =\; x$ ✓.

$f(x)=x^2-2f(x)\; =\; x^2\; -\; 2$ is not one-to-one on $\mathbb{R}\backslash mathbb\{R\}$. Restrict it to $x\ge 0x\; \backslash ge\; 0$ and find the inverse.

Solution On $[0,+\mathrm{\infty })[0,\; +\backslash infty)$, $ff$ is increasing, so one-to-one, with range $[-2,+\mathrm{\infty })[-2,\; +\backslash infty)$. $y=x^2-2⟺x^2=y+2⟺x=\sqrt{y+2}y\; =\; x^2\; -\; 2\; \backslash iff\; x^2\; =\; y\; +\; 2\; \backslash iff\; x\; =\; \backslash sqrt\{y\; +\; 2\}$ (the positive root, because $x\ge 0x\; \backslash ge\; 0$). So $f^{-1}(x)=\sqrt{x+2}f^\{-1\}(x)\; =\; \backslash sqrt\{x\; +\; 2\}$, $x\ge -2x\; \backslash ge\; -2$.