What is the domain of f(x)=(9x(x-4))/(6x^2-41x-7) ?

The domain of f(x)=(9x(x-4))/(6x^2-41x-7) is x 7

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domain of f(x)=(9x(x-4))/(6x^2-41x-7)

domain

f (x) = \frac{9 x ( x - 4 )}{6 x ^{2} - 41 x - 7}

x < - \frac{1}{6} or - \frac{1}{6} < x < 7 or x > 7

Interval Notation

(- \infty, - \frac{1}{6}) \cup (- \frac{1}{6}, 7) \cup (7, \infty)

Solution steps

Find undefined (singularity) points:

x = 7, x = - \frac{1}{6}

The function domain

x < - \frac{1}{6} or - \frac{1}{6} < x < 7 or x > 7

f (x) = \frac{( x + 1 )}{x ^{2} + 5 x + 6}

f (x) = 8^{x}

4 sin (\frac{1}{5} π x - π) + 3

(0, 0), (2 b, 0)

What is the domain of f(x)=(9x(x-4))/(6x^2-41x-7) ?
The domain of f(x)=(9x(x-4))/(6x^2-41x-7) is x<-1/6 \lor-1/6 <x<7\lor x>7