Answer to Question #243512 in Algebra for Trixie Saavedra

Question

Find the domain and range of the following rational function. Use any notation. 1. f(x) = 2

𝑥+1

2. f(x) = 3𝑥 𝑥+3

3. f(x) = 3−𝑥 𝑥−7

4. f(x) = 2+𝑥 𝑥

5. f(x) = (𝑥+1) 𝑥2−1

Answer 1

1. $f(x)=\dfrac{2}{x+1}$

$x+1\not=0=>x\not=-1$

$Domain:(-\infin, -1)\cup (-1, \infin)$

$Range:(-\infin, 0)\cup (0, \infin)$

2. $f(x)=\dfrac{3x}{x+3}$

$x+3\not=0=>x\not=-3$

$f(x)=\dfrac{3x}{x+3}=\dfrac{3x+9-9}{x+3}=3-\dfrac{9}{x+3}$

$Domain:(-\infin, -3)\cup (-3, \infin)$

$Range:(-\infin, 3)\cup (3, \infin)$

3. $f(x)=\dfrac{3-x}{x-7}$

$x-7\not=0=>x\not=7$

$f(x)=\dfrac{3-x}{x-7}=\dfrac{3-(x-7)-7}{x-7}=-1-\dfrac{4}{x-7}$

$Domain:(-\infin, 7)\cup (7, \infin)$

$Range:(-\infin, -1)\cup (-1, \infin)$

4. $f(x)=\dfrac{2+x}{x}$

$x\not=0$

$f(x)=\dfrac{2+x}{x}=1+\dfrac{2}{x}$ $Domain:(-\infin, 0)\cup (0, \infin)$

$Range:(-\infin, 1)\cup (1, \infin)$

5. $f(x)=\dfrac{x+1}{x^2-1}$

$x^2-1\not=0=>x\not=-1, x\not=1$

$f(x)=\dfrac{x+1}{x^2-1}=\dfrac{x+1}{(x+1)(x-1)}=-\dfrac{1}{x-1}, x\not=\pm1$

$Domain:(-\infin,-1)\cup (-1,1)\cup (1, \infin)$

$Range:(-\infin, 0)\cup (0, \infin)$