What is the range of y=(x+1)/(x-3) ?

The range of y=(x+1)/(x-3) is f(x) 1

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range of y=(x+1)/(x-3)

range

y = \frac{x + 1}{x - 3}

f (x) < 1 or f (x) > 1

Interval Notation

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

Solution steps

The function range is the combined domain of the inverse functions

Find the inverse function(s) of:

\frac{x + 1}{x - 3}

Inverse of

\frac{x + 1}{x - 3} : \frac{1 + 3 x}{x - 1}

\frac{1 + 3 x}{x - 1}

Find the domain of each inverse function

Domain of

\frac{1 + 3 x}{x - 1} : x < 1 or x > 1

Combine the intervals

f (x) < 1 or f (x) > 1

f (x) = \frac{x}{x ^{2} - 3}

f (x) = 2^{x + 3}

\frac{3 x ^{2} - 12 x + 13}{x ^{2} - 4 x + 4}

f (x) = 8 x^{3} - 12 x^{2} - 48 x

f (x) = x + x + 6

What is the range of y=(x+1)/(x-3) ?
The range of y=(x+1)/(x-3) is f(x)<1\lor f(x)>1