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

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

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

range

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

- 1 \leq f (x) < 1

Interval Notation

[- 1, 1)

Solution steps

Rewrite as

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

Multiply both sides by

x^{2} + 1

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

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

Discriminant

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

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

Check if the range interval endpoints are included

y = - 1 :

Included

y = 1 :

Excluded

Therefore the range is

- 1 \leq f (x) < 1

f (x) = \frac{( x - 9 )}{2}

(6, 3), (- 3, 4)

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

f (x) = - \frac{7 x}{6 x - 5}

f (x) = ln (x^{2} - 14 x)

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