What is the range of f(x)= 1/(x+3)+2 ?

The range of f(x)= 1/(x+3)+2 is f(x) 2

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range of f(x)= 1/(x+3)+2

range

f (x) = \frac{1}{x + 3} + 2

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

Interval Notation

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

Solution steps

The function range is the combined domain of the inverse functions

Find the inverse function(s) of:

\frac{1}{x + 3} + 2

Inverse of

\frac{1}{x + 3} + 2 : \frac{- 3 x + 7}{x - 2}

\frac{- 3 x + 7}{x - 2}

Find the domain of each inverse function

Domain of

\frac{- 3 x + 7}{x - 2} : x < 2 or x > 2

Combine the intervals

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

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

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

(5, - 14), (12, - 9)

y = x^{4} - 16 x^{2}

f (x) = \frac{1}{z - 1}

What is the range of f(x)= 1/(x+3)+2 ?
The range of f(x)= 1/(x+3)+2 is f(x)<2\lor f(x)>2