What is the range of (x^2)/(-2+x) ?

The range of (x^2)/(-2+x) is f(x) = 8

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

range of (x^2)/(-2+x)

range

\frac{x ^{2}}{- 2 + x}

f (x) \leq 0 or f (x) \geq 8

Interval Notation

(- \infty, 0] \cup [8, \infty)

Solution steps

Rewrite as

\frac{x ^{2}}{- 2 + x} = y

Multiply both sides by

- 2 + x

x^{2} = y (- 2 + x)

The range is the set of y for which the discriminant is greater or equal to zero

Discriminant

x^{2} = y (- 2 + x) : y^{2} - 8 y

y^{2} - 8 y \geq 0 : y \leq 0 or y \geq 8

Check if the range interval endpoints are included

y = 0 :

Included

y = 8 :

Included

Therefore the range is

f (x) \leq 0 or f (x) \geq 8

f (x) = \frac{x}{x + 8}

y = 2 x + 4 (4.4)

f (x) = 4 sec (6 x - 2 π) - 12

f (x) = (x + 2)^{2} - 1

lo g_{2} (x - 4)

What is the range of (x^2)/(-2+x) ?
The range of (x^2)/(-2+x) is f(x)<= 0\lor f(x)>= 8