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

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

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

range

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

f (x) < - 2 or f (x) > - 2

Interval Notation

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

Solution steps

Find the minimum and maximum value in each defined interval and unite the results

Domain of

\frac{x ^{2} - 1}{x + 1} : x < - 1 or x > - 1

Simplify

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

Extreme Points of

x - 1 :

None

Find the range for the interval

- \infty < x < - 1 : - \infty < f (x) < - 2

Find the range for the interval

- 1 < x < \infty : - 2 < f (x) < \infty

Combine the ranges of all domain intervals to obtain the function range

f (x) < - 2 or f (x) > - 2

f (x) < - 2 or f (x) > - 2

f (x) = \frac{x + 2}{x + 10}

f (x) = \frac{- x ^{2} - 4 x + 5}{4 x - 4}

f (x) = \frac{x - 4}{x - 6}

f (x) = x^{4} + x

2 x + 3

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