What is the range of f(x)= 1/(x^2-1) ?

The range of f(x)= 1/(x^2-1) is f(x) 0

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

range of f(x)= 1/(x^2-1)

range

f (x) = \frac{1}{x ^{2} - 1}

f (x) \leq - 1 or f (x) > 0

Interval Notation

(- \infty, - 1] \cup (0, \infty)

Solution steps

Rewrite as

\frac{1}{x ^{2} - 1} = y

Multiply both sides by

x^{2} - 1

1 = y (x^{2} - 1)

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

Discriminant

1 = y (x^{2} - 1) : 4 y^{2} + 4 y

4 y^{2} + 4 y \geq 0 : y \leq - 1 or y \geq 0

Check if the range interval endpoints are included

y = - 1 :

Included

y = 0 :

Excluded

Therefore the range is

f (x) \leq - 1 or f (x) > 0

f (x) = \frac{1}{x ^{6}}

8 x^{3} + 3 x

(4, - 1), (- 2, - 5)

x^{\frac{2}{3}} - 4

f (x) = 0

What is the range of f(x)= 1/(x^2-1) ?
The range of f(x)= 1/(x^2-1) is f(x)<=-1\lor f(x)>0