What is the domain of f(x)= 1/(\frac{4){x-3}-2} ?

The domain of f(x)= 1/(\frac{4){x-3}-2} is x 5

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domain of f(x)= 1/(\frac{4){x-3}-2}

domain

f (x) = \frac{1}{\frac{4}{x - 3} - 2}

x < 3 or 3 < x < 5 or x > 5

Interval Notation

(- \infty, 3) \cup (3, 5) \cup (5, \infty)

Solution steps

Find undefined (singularity) points:

x = 5, x = 3

The function domain

x < 3 or 3 < x < 5 or x > 5

f (x) = x^{4} - 4 x^{3} + 10

\frac{x ^{2} - 2 x - 8}{x ^{2} + x - 2}

f (x) = \frac{6 x - 9}{x + 5}

x + 6 - 9

f (x) = \frac{3 ( x + 7 ) ( x - 2 )}{( x - 3 )}

What is the domain of f(x)= 1/(\frac{4){x-3}-2} ?
The domain of f(x)= 1/(\frac{4){x-3}-2} is x<3\lor 3<x<5\lor x>5