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

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

range

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

f (x) > 0

Interval Notation

(0, \infty)

Solution steps

Rewrite as

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

Multiply both sides by

(1 - x)^{2}

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

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

Discriminant

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

4 y \geq 0 : y \geq 0

Check if the range interval endpoints are included

y = 0 :

Excluded

Therefore the range is

f (x) > 0

f (x) = \frac{x + 1}{2}

f (x) = \frac{1}{x + 9}

1 + x - 1

f (x) = \frac{6 e ^{x}}{e ^{x} - 7}

f (x) = - (x + 5)^{2} + 6

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