What is the domain of f(x)=1+tan^2(x) ?

The domain of f(x)=1+tan^2(x) is pin<= x< pi/2+pin\lor pi/2+pin<x<pi+pin

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

domain of f(x)=1+tan^2(x)

domain

f (x) = 1 + tan^{2} (x)

πn \leq x < \frac{π}{2} + πn or \frac{π}{2} + πn < x < π + πn

Interval Notation

[πn, \frac{π}{2} + πn) \cup (\frac{π}{2} + πn, π + πn)

Solution steps

Take the denominator(s) of

1 + tan^{2} (x)

and compare to zero:

x = \frac{π}{2} + πn

Periodicity of

1 + tan^{2} (x) : π

Combine period:

πn \leq x < π + πn

and undefined points.

The domain is

πn \leq x < \frac{π}{2} + πn or \frac{π}{2} + πn < x < π + πn

a sy m pt o t es f (x) = 3^{x} + 1

s l o p e in t erce pt 3 x - 6 y = 6

a sy m pt o t es \frac{2 x ^{2}}{( x + 2 ) ( x - 3 )}

in v erse f (x) = (x - 2)^{2}, x \geq 2

d o main \frac{x}{x ^{2} + 16}

What is the domain of f(x)=1+tan^2(x) ?
The domain of f(x)=1+tan^2(x) is pin<= x< pi/2+pin\lor pi/2+pin<x<pi+pin