How do you find the domain, range, and asymptote for $y = 1 + cot (3 x + \frac{π}{2})$ $y = 1 + cot ( 3x + pi/2 )$ ?

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Answer 1 · 2016-11-22T14:36:06

Graph reveals all details. Noe that $π = 3.14$ $pi=3.14$ , nearly, See explanation, for numerical facts.

graph{y-1+tan( 3x)=0 [-10, 10, -5, 5]}

$y = 1 + cot (3 x + \frac{π}{2}) = 1 - tan 3 x$ $y=1+cot(3x+pi/2)=1-tan 3x$

The period for the graph is $\frac{π}{3}$ $pi/3$ . is

y is infinitely discontinuous at #x= k/6pi+pi/3), k = 0, +-1, +-2, +-3

The piecewise domain:

$x \in (\frac{k}{6} π, \frac{k}{6} π + \frac{π}{3}), k = 0, \pm 1, \pm 2, \pm 3, . .$ $x in (k/6pi, k/6pi+pi/3), k = 0, +-1, +-2, +-3, ..$ .

Range: $y \in (- \infty, \infty)$ $y in (-oo, oo)$ .+-3, ...#

As $x to (k/6pi, k/6pi+pi/3), k = 0, +-1, +-2, +-3, ...y to +-oo$

Asymptotes: $x=k/6pi+pi/3, k = 0, +-1, +-2, +-3, ...$

Have a look at the points inflexion aligned upon y =1,

wherein $x=.k/3pi, k = 0, +-1, +-2, +-3, ...$

How do you find the domain, range, and asymptote for y = 1 + cot ( 3x + pi/2 )y=1+cot(3x+π2)y = 1 + cot ( 3x + pi/2 )?