What is the range of f(x)=(2x^2-3)/(x^2-1) ?

The range of f(x)=(2x^2-3)/(x^2-1) is f(x) = 3

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range of f(x)=(2x^2-3)/(x^2-1)

range

f (x) = \frac{2 x ^{2} - 3}{x ^{2} - 1}

f (x) < 2 or f (x) \geq 3

Interval Notation

(- \infty, 2) \cup [3, \infty)

Solution steps

Rewrite as

\frac{2 x ^{2} - 3}{x ^{2} - 1} = y

Multiply both sides by

x^{2} - 1

2 x^{2} - 3 = y (x^{2} - 1)

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

Discriminant

2 x^{2} - 3 = y (x^{2} - 1) : 4 y^{2} - 20 y + 24

4 y^{2} - 20 y + 24 \geq 0 : y \leq 2 or y \geq 3

Check if the range interval endpoints are included

y = 2 :

Excluded

y = 3 :

Included

Therefore the range is

f (x) < 2 or f (x) \geq 3

f (x) = - \frac{6}{x}

f (x) = \frac{x - 3}{4}

y = \frac{- 5}{1.25} (x - 8.5) + 7

3 x^{3} - 4 x^{2} + 3 x - 2

- x + y = - 2

What is the range of f(x)=(2x^2-3)/(x^2-1) ?
The range of f(x)=(2x^2-3)/(x^2-1) is f(x)<2\lor f(x)>= 3