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

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

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

range

f (x) = \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

in t erce pt s y = 3 x^{2} + 8 x - 3

in v erse f (x) = 4 x^{2} + 3

p er p e n d i c u l a r 5, - 6

d o main f (x) = \frac{3}{x + 5}

s l o p e 8 sin (\frac{π}{6})

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