What is the domain of (sqrt(19-x))/(x-8)+ln(3x+48) ?

The domain of (sqrt(19-x))/(x-8)+ln(3x+48) is -16<x<8\lor 8<x<= 19

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domain of (sqrt(19-x))/(x-8)+ln(3x+48)

domain

\frac{19 - x}{x - 8} + ln (3 x + 48)

- 16 < x < 8 or 8 < x \leq 19

Interval Notation

(- 16, 8) \cup (8, 19]

Solution steps

Find non-negative values for radicals:

x \leq 19

Find positive values for logs:

x > - 16

Find undefined (singularity) points:

x = 8

Combine real regions and undefined points for final function domain

- 16 < x < 8 or 8 < x \leq 19

(11 x + 2 x^{2}) - (- 7 - 3 x^{2} + 4)

\frac{21}{( x + 4 ) x}

f (x) = 2 e^{- x^{2} - x^{2}} + 2 cos (x) + 3

\frac{w ^{2} + w - 12}{w ^{2} - 81}

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

What is the domain of (sqrt(19-x))/(x-8)+ln(3x+48) ?
The domain of (sqrt(19-x))/(x-8)+ln(3x+48) is -16<x<8\lor 8<x<= 19