Question

Find the domain and range of the following rational function. Use qny 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) / x^2-1

Answer 1

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

Here x+1 $\not =$ 0 // otherwise function will not be defined

So x $\not =$ - 1

Domain ($-\infty,-1)$ $\bigcup$ $(-1 , \infty)$

Range $(-\infty ,0) \bigcup (0,\infty)$

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

Here x+3 $\not =$ 0 // otherwise function will not be defined

So x $\not =$ -3

Domain $(-\infty ,-3) \bigcup (-3 ,\infty)$

f(x) = $\frac{3x +9 -9}{x+3}$ = $\frac{3(x+3) - 9}{x+3}$ = 3 - $\frac{9}{x+3}$ ----------------------- (1)

From equation (1) we can understand the value will never reach 3.

So Range $(-\infty ,3) \bigcup (3, \infty)$

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

Here x-7 $\not =$ 0 // otherwise function will not be defined

So x $\not =$ 7

Domain $(-\infty ,7) \bigcup (7,\infty)$

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

Range $(-\infty ,-1) \bigcup (-1,\infty)$

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

Here x $\not =$ 0

Domain $(-\infty ,0) \bigcup (0,\infty)$

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

Range $(-\infty,1) \bigcup (1,\infty)$

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

Here x² - 1 $\not =$ 0

So x $\not = 1$ and x $\not = -1$

Domain $(-\infty,-1) \bigcup (-1,1) \bigcup (1,\infty)$

Range $(-\infty ,0) \bigcup (0,\infty)$