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

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

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

range

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

- \infty < f (x) < \infty

Interval Notation

(- \infty, \infty)

Solution steps

Rewrite as

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

Multiply both sides by

x^{2} - 1

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

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

Discriminant

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

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

True for all

y \in R

Therefore the range is

- \infty < f (x) < \infty

f (x) = \frac{( 2 x )}{( x + 5 )}

2 x - 5 y - 6 = 0

f (t) = \frac{2}{t ^{2} - 16}

8 (q - 13)

f (x) = 3 - 2 x - x^{2}

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