What is the domain of (-7(4+x))/((2^x-8)(-1-3x)^2) ?

The domain of (-7(4+x))/((2^x-8)(-1-3x)^2) is x 3

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domain of (-7(4+x))/((2^x-8)(-1-3x)^2)

domain

\frac{- 7 ( 4 + x )}{( 2 ^{x} - 8 ) ( - 1 - 3 x ) ^{2}}

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

Interval Notation

(- \infty, - \frac{1}{3}) \cup (- \frac{1}{3}, 3) \cup (3, \infty)

Solution steps

Find undefined (singularity) points:

x = 3, x = - \frac{1}{3}

The function domain

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

y = 20 - 3 x

ln (7 - 6 x^{2})

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

(5, - 2), (1, 2)

f (x) = \frac{x + 18}{x - 17}

What is the domain of (-7(4+x))/((2^x-8)(-1-3x)^2) ?
The domain of (-7(4+x))/((2^x-8)(-1-3x)^2) is x<-1/3 \lor-1/3 <x<3\lor x>3