What is the domain of f(t)=sec((pit)/4) ?

The domain of f(t)=sec((pit)/4) is 8n<= t<2+8n\lor 2+8n<t<6+8n\lor 6+8n<t<8+8n

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domain of f(t)=sec((pit)/4)

domain

f (t) = sec (\frac{π t}{4})

8 n \leq t < 2 + 8 n or 2 + 8 n < t < 6 + 8 n or 6 + 8 n < t < 8 + 8 n

Interval Notation

[8 n, 2 + 8 n) \cup (2 + 8 n, 6 + 8 n) \cup (6 + 8 n, 8 + 8 n)

Solution steps

Take the denominator(s) of

sec (\frac{π t}{4})

and compare to zero:

t = 2 + 8 n, t = 6 + 8 n

Periodicity of

sec (\frac{π t}{4}) : 8

Combine period:

8 n \leq t < 8 + 8 n

and undefined points.

The domain is

8 n \leq t < 2 + 8 n or 2 + 8 n < t < 6 + 8 n or 6 + 8 n < t < 8 + 8 n

f (x) = 4 x^{2} - 16

lo g_{\frac{1}{4}} (x)

\frac{4 x + 3}{2 x - 1}

f (x) = \frac{x}{x ^{2} - 5}

f (x) = - 2 x^{2} + 12 x + 5

What is the domain of f(t)=sec((pit)/4) ?
The domain of f(t)=sec((pit)/4) is 8n<= t<2+8n\lor 2+8n<t<6+8n\lor 6+8n<t<8+8n