Domain and range of polynomial, rational and irrational functions

Find the domain of (a) $f(x)=\sqrt{x^2-x-6}f(x)\; =\; \backslash sqrt\{x^2\; -\; x\; -\; 6\}$ (b) $g(x)=\frac{1}{\sqrt{4-x}}+\sqrt{x+1}g(x)\; =\; \backslash frac\{1\}\{\backslash sqrt\{4\; -\; x\}\}\; +\; \backslash sqrt\{x\; +\; 1\}$ (c) $h(x)=\sqrt{\frac{x-1}{x+2}}h(x)\; =\; \backslash sqrt\{\backslash frac\{x\; -\; 1\}\{x\; +\; 2\}\}$.

Solution

1. $x^2-x-6=(x-3)(x+2)\ge 0x^2\; -\; x\; -\; 6\; =\; (x\; -\; 3)(x\; +\; 2)\; \backslash ge\; 0$. The trinomial is non-negative outside its roots: $\mathrm{Dom}f=(-\mathrm{\infty },-2]\cup [3,+\mathrm{\infty })\backslash operatorname\{Dom\}\; f\; =\; (-\backslash infty,\; -2]\; \backslash cup\; [3,\; +\backslash infty)$.
2. $4-x>04\; -\; x\; \backslash gt\; 0$ (root in a denominator) gives $x<4x\; \backslash lt\; 4$; $x+1\ge 0x\; +\; 1\; \backslash ge\; 0$ gives $x\ge -1x\; \backslash ge\; -1$. Both: $\mathrm{Dom}g=[-1,4)\backslash operatorname\{Dom\}\; g\; =\; [-1,\; 4)$.
3. $\frac{x-1}{x+2}\ge 0\backslash frac\{x\; -\; 1\}\{x\; +\; 2\}\; \backslash ge\; 0$ with $x\mathrm{\ne }-2x\; \backslash ne\; -2$. A sign table gives $\mathrm{Dom}h=(-\mathrm{\infty },-2)\cup [1,+\mathrm{\infty })\backslash operatorname\{Dom\}\; h\; =\; (-\backslash infty,\; -2)\; \backslash cup\; [1,\; +\backslash infty)$.

Find the range of (a) $f(x)=x^2-4x+1f(x)\; =\; x^2\; -\; 4x\; +\; 1$ (b) $g(x)=-2x^2+8xg(x)\; =\; -2x^2\; +\; 8x$.

Solution

1. $f(x)=(x-2{)}^2-3\ge -3f(x)\; =\; (x\; -\; 2)^2\; -\; 3\; \backslash ge\; -3$, with equality at $x=2x\; =\; 2$. Range $[-3,+\mathrm{\infty })[-3,\; +\backslash infty)$.
2. $g(x)=-2(x^2-4x)=-2(x-2{)}^2+8\le 8g(x)\; =\; -2(x^2\; -\; 4x)\; =\; -2(x\; -\; 2)^2\; +\; 8\; \backslash le\; 8$, with equality at $x=2x\; =\; 2$. Range $(-\mathrm{\infty },8](-\backslash infty,\; 8]$.

Find the domain and range of $f(x)=\frac{2x+1}{x-1}f(x)\; =\; \backslash frac\{2x\; +\; 1\}\{x\; -\; 1\}$.

Solution Domain: $\mathbb{R}\setminus \{1\}\backslash mathbb\{R\}\; \backslash setminus\; \backslash \{1\backslash \}$. For the range, solve $y=\frac{2x+1}{x-1}y\; =\; \backslash frac\{2x\; +\; 1\}\{x\; -\; 1\}$ for $xx$:

For $y\mathrm{\ne }2y\; \backslash ne\; 2$ this gives an $xx$, and that $xx$ is never 1 (since $\frac{y+1}{y-2}=1\backslash frac\{y\; +\; 1\}\{y\; -\; 2\}\; =\; 1$ would need $y+1=y-2y\; +\; 1\; =\; y\; -\; 2$). For $y=2y\; =\; 2$ the equation $0x=30x\; =\; 3$ has no solution. Range: $\mathbb{R}\setminus \{2\}\backslash mathbb\{R\}\; \backslash setminus\; \backslash \{2\backslash \}$.

Find the domain and range of (a) $f(x)=\sqrt{4-x^2}f(x)\; =\; \backslash sqrt\{4\; -\; x^2\}$ (b) $g(x)=1+\sqrt{x-3}g(x)\; =\; 1\; +\; \backslash sqrt\{x\; -\; 3\}$.

Solution

1. $4-x^2\ge 0⟺-2\le x\le 24\; -\; x^2\; \backslash ge\; 0\; \backslash iff\; -2\; \backslash le\; x\; \backslash le\; 2$. On this interval $4-x^{2}4\; -\; x^2$ takes every value from 0 (at $x=\pm 2x\; =\; \backslash pm\; 2$) to 4 (at $x=0x\; =\; 0$), so $\sqrt{4-x^2}\backslash sqrt\{4\; -\; x^2\}$ takes every value from 0 to 2. Domain $[-2,2][-2,\; 2]$, range $[0,2][0,\; 2]$. Squaring $y=\sqrt{4-x^2}y\; =\; \backslash sqrt\{4\; -\; x^2\}$ gives $x^2+y^2=4x^2\; +\; y^2\; =\; 4$ with $y\ge 0y\; \backslash ge\; 0$: the graph is the upper half of a circle of radius 2.
2. Domain $[3,+\mathrm{\infty })[3,\; +\backslash infty)$. Since $\sqrt{x-3}\backslash sqrt\{x\; -\; 3\}$ takes every value $\ge 0\backslash ge\; 0$, the range is $[1,+\mathrm{\infty })[1,\; +\backslash infty)$.