Can someone help me find the domain and range of y=3csc(3x+pi)-2 ?
idk because Im in 7th grade...
@ganeshie8 Can you help me with this math problem?
@e.mccormick @thomaster Can you help me with this math problem? (AP Calculus, Trig review)
I know that the domain is supposed to be kpi/3, but I don't know how to get that.
@e.mccormick
@thomaster
@amistre64 @ganeshie8 Can you help me?
the csc function rides on the humps of which function?
whenever the function is rides on goes zero, the inverse hits a divide by zero and we eliminate those from the domain
the 'parent' functions range llives between a min and max, the inverse hits everything else everything else
in other words, exploring the parent function will help us determine this one
@amistre64 so how do I do that to find the domain?
what is the 'parent' function that csc is an inverse of?
y=sin x ?
inverse is a bad name for it in trig .... reciprocal and yes, sin of whtever argument they used
now, there is only one real number that doesnt have a good reciprocal: 0 if we flip it we get n/0 soo ... we need to exclude every argument that makes sin go zero
How do I find what makes sin zero?
um, that is going to be one of those basic trig questions that you should have in memory already
sin(u) = 0 then solve the argument they give you to equate it to u
What is u?
u is an argument that we can use that is not x ....
I feel like I should know all of this, but I don't remember learning it. The textbook doesn't say anything about it either.
when does sin equal zero? you need to know this, ill correct you if need be but i cant just tell you outright
You mean on the unit circle?
i cant draw at the moment or id demonstrate a unit circle
yeah
Everyone else online was saying you have to solve for sin (3x+ pi) = k pi or something like that so maybe that's confusing me but I'm not sure.
itll be more like, find when sin(u) = 0 then u = 3x+pi will solve for x (u - pi)/3 = x
I'm pretty sure its zero at 0 and pi
0 and pi are good, but aslo 2pi, 3pi, 4pi .... every integer multiple of pi makes sin go zero
so, lets say u = k pi, such that k is an integer
okay
so, solving for x, as i already did (k pi-pi)/3 = x has to be excluded from the set of Real numbers and that defines our domain
(k-1)pi/3 but since k is an integer, then k-1 is also any integer lets say every k pi/3 for brevity
\[D=\{x \in R:x\ne k\frac{\pi}{3}, k\in Z\}\]
now the range is simple enough, at least for me since i already know my trig :) its the center line, plus/minus the amplitude
(-infinity, -5) U (1, infinity) ?
include the -5 and 1
sin=csc at the humps
What do you mean by include
your notation, when we want to omit an endpoint we close it with ) when we want to include it, it EQUALS it, we draw an equal sign connected by a bar: ]
since infinity can never be equated, you use ( and ) which is fine, but -5 and 1 are PART of the range and need to be included since we can equal -5 or 1
Oh okay so just use brackets instead of parentheses?
for the -5 and 1 yes
ok thank you. There's another problem like this but it uses tangent instead of csc. Do you think you could help me out on that as well?
tangent is a common trig function, its built out of sin and cos: tan = sin/cos so when do you think we would omit a part of the domain?
At pi/2 and 3pi/2?
That's when cot is zero I believe
odd integer multiples of pi/2 yep
the equation is y = -3 tan (2x + pi) + 2
Not sure what to do next..
oops I meant y = -3 tan (3x + pi) + 2
3x + pi = kpi/2 ?
I'll take that as a yes..
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