What is the domain of f(x)=sec(5x+pi) ?

The domain of f(x)=sec(5x+pi) is

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domain of f(x)=sec(5x+pi)

domain

f (x) = sec (5 x + π)

\frac{2 π}{5} n \leq x < \frac{π}{10} + \frac{2 π}{5} n or \frac{π}{10} + \frac{2 π}{5} n < x < \frac{3 π}{10} + \frac{2 π}{5} n or \frac{3 π}{10} + \frac{2 π}{5} n < x < \frac{2 π}{5} + \frac{2 π}{5} n

Interval Notation

[\frac{2 π}{5} n, \frac{π}{10} + \frac{2 π}{5} n) \cup (\frac{π}{10} + \frac{2 π}{5} n, \frac{3 π}{10} + \frac{2 π}{5} n) \cup (\frac{3 π}{10} + \frac{2 π}{5} n, \frac{2 π}{5} + \frac{2 π}{5} n)

Solution steps

Take the denominator(s) of

sec (5 x + π)

and compare to zero:

x = \frac{3 π}{10} + \frac{2 π}{5} n, x = \frac{π}{10} + \frac{2 π}{5} n

Periodicity of

sec (5 x + π) : \frac{2 π}{5}

Combine period:

\frac{2 π}{5} n \leq x < \frac{2 π}{5} + \frac{2 π}{5} n

and undefined points.

The domain is

\frac{2 π}{5} n \leq x < \frac{π}{10} + \frac{2 π}{5} n or \frac{π}{10} + \frac{2 π}{5} n < x < \frac{3 π}{10} + \frac{2 π}{5} n or \frac{3 π}{10} + \frac{2 π}{5} n < x < \frac{2 π}{5} + \frac{2 π}{5} n

f (x) = \frac{∣ x - 1 ∣}{x - 1} x^{2}

f (x) = \frac{x ^{2}}{9} + \frac{( y - 5 ) ^{2}}{25} = 1

+ (- 3) x + 3

f (x) = (x^{2} - 4) (x + 6)

x (160 - x)