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

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

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

range

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

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

Interval Notation

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

Solution steps

Rewrite as

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

Multiply both sides by

x^{2} - 1

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

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

Discriminant

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

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

Check if the range interval endpoints are included

y = 0 :

Included

y = 1 :

Excluded

Therefore the range is

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

y = - 6 x^{3} + 2 x

f (x) = x^{2} ∣ x ∣ + 3

f (x) = 2 x^{2} + 8 x

\frac{4}{x - 3}

f (x) = \frac{( 4 x )}{( 9 x - 1 )}

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