How do you find the domain and range of $h(x)= 10/(x^2-2x)$ ?

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Answer 1 · 2015-07-22T10:14:39

Analyse behaviour of the denominator and hence $h(x)$ to find:

domain $= (-oo, 0) uu (0, 2) uu (2, oo)$

and

range $= (-oo, -10] uu (0, oo)$

$h(x) = 10/(x^2-2x) = 10/(x(x-2)) = 10/((x-1)^2-1)$

The denominator is zero when $x=0$ or $x=2$ , so $h(x)$ is undefined for those values of $x$ .

When $x in (0, 2)$ , $x^2-2x in [-1,0)$ ,

so $h(x) in (-oo, -10]$

When $x in (-oo, 0) uu (2, oo)$ , $x^2-2x in (0, oo)$ ,

so $h(x) in (0, oo)$

How do you find the domain and range of h(x)= 10/(x^2-2x)h(x)= 10/(x^2-2x)?