What is the range of y= 1/(x+2) ?

The range of y= 1/(x+2) is f(x) 0

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range of y= 1/(x+2)

range

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

f (x) < 0 or f (x) > 0

Interval Notation

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

Solution steps

The function range is the combined domain of the inverse functions

Find the inverse function(s) of:

\frac{1}{x + 2}

Inverse of

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

\frac{1}{x} - 2

Find the domain of each inverse function

Domain of

\frac{1}{x} - 2 : x < 0 or x > 0

Combine the intervals

f (x) < 0 or f (x) > 0

4 x^{3} - 5

\frac{x + 3}{x - 2}

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

\frac{1}{x - 4} + 2

f (x) = \frac{2 x ^{2} - 5 x}{2 x + 3}