What is the domain and range for $f (x) = \frac{x}{x^{2} - 5 x}$ $f(x)= x/(x^2-5x)$ ?

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What is the domain and range for $f (x) = \frac{x}{x^{2} - 5 x}$ $f(x)= x/(x^2-5x)$ ?

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Answer 1 · 2015-09-20T20:57:25

The domain of $f (x)$ $f(x)$ is $(- \infty, 0) \cup (0, 5) \cup (5, \infty)$ $(-oo, 0) uu (0, 5) uu (5, oo)$ and

the range of $f (x)$ $f(x)$ is $(- \infty, - \frac{1}{5}) \cup (- \frac{1}{5}, 0) \cup (0, \infty)$ $(-oo, -1/5) uu (-1/5, 0) uu (0, oo)$ .

Explanation:

$f (x) = \frac{x}{x^{2} - 5 x} = \frac{x}{x (x - 5)} = \frac{1}{x - 5}$ $f(x) = x/(x^2-5x) = x/(x(x-5)) = 1/(x-5)$ with exclusion $x \neq 0$ $x != 0$

The denominator of $f (x)$ $f(x)$ is zero when $x = 0$ $x=0$ or $x = 5$ $x=5$ .

Let $y = f (x) = \frac{1}{x - 5}$ $y = f(x) = 1/(x-5)$ . Then $x = \frac{1}{y} + 5$ $x = 1/y + 5$ .

Therefore $y = 0$ $y = 0$ is an excluded value. Also $y = - \frac{1}{5}$ $y = -1/5$ is an excluded value, since it would result in $x = 0$ $x = 0$ , which is an excluded value.

So the domain of $f (x)$ $f(x)$ is $(- \infty, 0) \cup (0, 5) \cup (5, \infty)$ $(-oo, 0) uu (0, 5) uu (5, oo)$ and

the range of $f (x)$ $f(x)$ is $(- \infty, - \frac{1}{5}) \cup (- \frac{1}{5}, 0) \cup (0, \infty)$ $(-oo, -1/5) uu (-1/5, 0) uu (0, oo)$ .

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What is the domain and range for $f (x) = \frac{x}{x^{2} - 5 x}$ $f(x)= x/(x^2-5x)$ ?

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Explanation:

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