What is the domain and range of $h(x)= 10/(x^2-2x)$ ?

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Answer 1 · 2016-06-19T07:04:26

Domain is $(-oo,0) uu (0,2)uu(2,+oo)$
Range is $(-oo,-40/9]uu(0,+oo)$

The domain is obtained by solving:

$x^2-2x!=0$

$x(x-2)!=0$

$x!=0 and x!=2$

You can find the range by calculating the inverse function

Let y=h(x)

so

$y=10/(x^2-3x)$

$yx^2-3xy-10=0$

$x=(3y+-sqrt(9y^2-4y(-10)))/(2y)$

you can find its domain by solving:

$9y^2+40y>=0 and y!=0$

$y(9y+40)>=0 and y!=0$

$y<=-40/9 or y>0$

What is the domain and range of h(x)= 10/(x^2-2x)h(x)= 10/(x^2-2x)?