What is the range of y=(-7)/(1x-3) ?

The range of y=(-7)/(1x-3) is f(x) 0

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range of y=(-7)/(1x-3)

range

y = \frac{- 7}{1 x - 3}

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{- 7}{1 x - 3}

Inverse of

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

- \frac{7}{x} + 3

Find the domain of each inverse function

Domain of

- \frac{7}{x} + 3 : x < 0 or x > 0

Combine the intervals

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

f (x) = \frac{60}{x ( x + 4 )}

\frac{5 ^{x}}{5 ^{x} + 3 ^{x}}

f (x) = \frac{4}{x + 3} + 2

\frac{e ^{x}}{x}

3 sin (x)

What is the range of y=(-7)/(1x-3) ?
The range of y=(-7)/(1x-3) is f(x)<0\lor f(x)>0