inverse graphs (1 Viewer)

[alez]

having major probs drawing the inverse graphs
like if theres something like 4sin^-1(2x-3)+5 i have absolutely no idea what to do
anyone have a good set of rules or something for that cuz we didnt really do anything on it in class
also domain/range for inverse graphs/questions
help

 

[vds700]

i think a good place to start would be to find the domain and range.

y = 4arcsin(2x - 3) + 5

we can write it as:
(y - 5)/4 = arcsin(2x - 3)

Domain is
-1 < 2x - 3 < 1
2 < 2x < 4
1 < x < 2

so x is between 1 and 2

range:
-pi/2 < (y - 5)/4 < pi/2
-2pi < y - 5 < 2pi
5 - 2pi < y < 5 + 2pi

all you have to do is draw a standard inverse sin graph with this domain and range.

NOTE:where i have typed the "<", i mean less than or equal to.

 

[lacklustre]

Hey guys, I thought I might post the questions here rather than create a new thread on the same topic.
I too am having difficulties sketching some inverse graphs:

1. y = 2sin^-1(1-x²)

2. y = 4cos^-1sqrt(1-x²)

3. y = cos^-1(cosx)


Need some help these are really driving me crazy!

 

[lacklustre]

omg guys. google.

http://www.walterzorn.com/grapher/grapher_e.htm

I know what the graphs look like (my book has them in the answers) but I don't know how to get there. That's the problem.

 

[3unitz]

I know what the graphs look like (my book has them in the answers) but I don't know how to get there. That's the problem.

sin^-1 (x) has domain -1 < x < 1

sin^-1 (0) = 0

sin^-1 (1) = pi/2

thats all you need to know

 

[3unitz]

1. y = 2sin^-1(1-x²)

y = 2 sin^-1(1 - x^2)

-1 < 1 - x^2 < 1

-2 < - x^2 < 0

2 > x^2 > 0

take 2 cases for +/- square root.

y = 2 sin^-1(1 - x^2)

sub y = 0 to find x intercepts

sub x = 0 to find y intercept

 

[lacklustre]

Ok ty for that. I think i get it.

For q 2, i am a little unsure how to find the domain. This is what i do:

-1 < sqrt(1-x²) < 1 (where "<" means less than or equal to)

I solve that but something is wrong because i get the domain to be:
0 < x < 0 , which is obviously wrong.

And for question 3 i am quite confused because i think it has something to do with absolute values.

 

[pLuvia]

Well firstly you've already violated one of the most fundamental mathematical laws

sqrt(x)>0 for x real>0
undefined otherwise (unless we talking in a complex space)

So the real domain is
0< sqrt(1-x²) <1
0< 1-x² <1
0< x < sqrt(1²)
0< x <|1|
-1< x <1

 

[lacklustre]

Well firstly you've already violated one of the most fundamental mathematical laws

sqrt(x)>0 for x real>0
undefined otherwise (unless we talking in a complex space)

So the real domain is
0< sqrt(1-x²) <1
0< 1-x² <1
0< x < sqrt(1²)
0< x <|1|
-1< x <1

Oh yeah! I have to learn to think more logically next time. Cheers.

 

[lacklustre]

Well firstly you've already violated one of the most fundamental mathematical laws

sqrt(x)>0 for x real>0
undefined otherwise (unless we talking in a complex space)

So the real domain is
0< sqrt(1-x²) <1
0< 1-x² <1
0< x < sqrt(1²)
0< x <|1|
-1< x <1

ok i don't get that part

 

[3unitz]

0 < x^2 < 1
-1 < x < 1

test out some values on your calculator and you'll see why