What is the domain of y=tan(4x-pi/4) ?

The domain of y=tan(4x-pi/4) is pi/4 n<= x<(3pi)/(16)+pi/4 n\lor (3pi)/(16)+pi/4 n<x< pi/4+pi/4 n

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domain of y=tan(4x-pi/4)

domain

y = tan (4 x - \frac{π}{4})

\frac{π}{4} n \leq x < \frac{3 π}{16} + \frac{π}{4} n or \frac{3 π}{16} + \frac{π}{4} n < x < \frac{π}{4} + \frac{π}{4} n

Interval Notation

[\frac{π}{4} n, \frac{3 π}{16} + \frac{π}{4} n) \cup (\frac{3 π}{16} + \frac{π}{4} n, \frac{π}{4} + \frac{π}{4} n)

Solution steps

Take the denominator(s) of

tan (4 x - \frac{π}{4})

and compare to zero:

x = \frac{3 π}{16} + \frac{π}{4} n

Periodicity of

tan (4 x - \frac{π}{4}) : \frac{π}{4}

Combine period:

\frac{π}{4} n \leq x < \frac{π}{4} + \frac{π}{4} n

and undefined points.

The domain is

\frac{π}{4} n \leq x < \frac{3 π}{16} + \frac{π}{4} n or \frac{3 π}{16} + \frac{π}{4} n < x < \frac{π}{4} + \frac{π}{4} n

x^{2} - 8 x + 25

F (x) = \frac{1}{x ^{2} + 25}

lo g_{4} (- 1 - 9 x)

y = \frac{1 + 6 x}{x - 1}

h (x) = (5 x - 25)^{- 2}

What is the domain of y=tan(4x-pi/4) ?
The domain of y=tan(4x-pi/4) is pi/4 n<= x<(3pi)/(16)+pi/4 n\lor (3pi)/(16)+pi/4 n<x< pi/4+pi/4 n