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

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

2 x^{3} - 5 x^{2} + 4 x + 2

x^{2} + x + 2

- x + \frac{3}{4} y = - 6

f (x) = - \frac{10}{x - 8}

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