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

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Answer 1 · 2015-09-20T20:57:25

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

the range of $f(x)$ is $(-oo, -1/5) uu (-1/5, 0) uu (0, oo)$ .

$f(x) = x/(x^2-5x) = x/(x(x-5)) = 1/(x-5)$ with exclusion $x != 0$

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

Let $y = f(x) = 1/(x-5)$ . Then $x = 1/y + 5$ .

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

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

the range of $f(x)$ is $(-oo, -1/5) uu (-1/5, 0) uu (0, oo)$ .

What is the domain and range for f(x)= x/(x^2-5x)f(x)= x/(x^2-5x)?