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

Question

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

1 Answer

Impact of this question

2712 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2016-06-19T07:04:26

Domain is $(- \infty, 0) \cup (0, 2) \cup (2, + \infty)$ $(-oo,0) uu (0,2)uu(2,+oo)$
Range is $(- \infty, - \frac{40}{9}] \cup (0, + \infty)$ $(-oo,-40/9]uu(0,+oo)$

Explanation:

The domain is obtained by solving:

$x^{2} - 2 x \neq 0$ $x^2-2x!=0$

$x (x - 2) \neq 0$ $x(x-2)!=0$

$x \neq 0 and x \neq 2$ $x!=0 and x!=2$

You can find the range by calculating the inverse function

Let y=h(x)

so

$y = \frac{10}{x^{2} - 3 x}$ $y=10/(x^2-3x)$

$y x^{2} - 3 x y - 10 = 0$ $yx^2-3xy-10=0$

$x = \frac{3 y \pm \sqrt{9 y^{2} - 4 y (- 10)}}{2 y}$ $x=(3y+-sqrt(9y^2-4y(-10)))/(2y)$

you can find its domain by solving:

$9 y^{2} + 40 y \geq 0 and y \neq 0$ $9y^2+40y>=0 and y!=0$

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

$y \leq - \frac{40}{9} or y > 0$ $y<=-40/9 or y>0$

Science

Math

Humanities

... and beyond

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

1 Answer

Explanation:

Related questions

Impact of this question

Science

Math

Humanities

... and beyond

What is the domain and range of h(x)=10x2−2xh(x)= 10/(x^2-2x)?

1 Answer

Explanation:

Related questions

Impact of this question

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