help (1 Viewer)

[iheartcookies1012]

how do u find the domain and range of y=sin(2cos^-1 x)?

 

[synthesisFR]

2arccos x has a range between 0 to 2pi, and since this is the angle that in in the sin(), so sin's range is restricted by -1 and 1 bc its a sin curve
the domain would be -1 to 1 as 2arccos x domain is limited to -1 to 1 so past that the sin wont exist teither

 

[iheartcookies1012]

thank you so much!!

 

[iheartcookies1012]

also how would you graph y=sin^-1 (sqrtx) and y=sin^-1 (x^2)?

 

[iheartcookies1012]

wait but for sin^-1(sqrtx) there is a restriction on range... its [0,pi/2] instead of [-pi/2, pi/2], is this becus the domain got restricted so the range also changed?

 

[synthesisFR]

wait but for sin^-1(sqrtx) there is a restriction on range... its not [0,pi/2] instead of [-pi/2, pi/2], is this becus the domain got restricted so the range also changed?

wait lmao my bad i must have swapped the two ill edit it
sorry for the confusion

 

[synthesisFR]

p.s this is ext math not adv just so u know

start with domain and range
for:
A) y=sin^-1 (sqrtx)
the inside exists for x>=0
domain :so the curve exists from [0,1]
range: would be [0,pi/2] because of sqrt x only existing for positive values the curve doesn't exist left of the x axis, cutting the range of at 0.


B) and y=sin^-1 (x^2)
domain is from -1 to 1 since there are not restrictions on x^2
range is from [0,pi/2], as ur squaring x, the curve will only take positive values as all the angles are positive in this domain. The part to the right of the y axis will be mirrored to the left side.

however, depending on how much detail u need u could change how much the curve dilates and stuff compared to a normal arcsinx graph.
- tho i dont think its required for u to consider this at all for inverse trig graphs

 

[synthesisFR]

can u read and see if that makes sense or works
im doing english rn and my draft is due at 11.59pm so i dont have much time to check : p

 

[iheartcookies1012]

can u read and see if that makes sense or works
im doing english rn and my draft is due at 11.59pm so i dont have much time to check : p

yep it does thanks !! gl w english (i hate it too)

 

[Average Boreduser]

wait but for sin^-1(sqrtx) there is a restriction on range... its [0,pi/2] instead of [-pi/2, pi/2], is this becus the domain got restricted so the range also changed?

What I'd recommend u do is let u=f(x) in y=sin^-1[f(x)] and using this information, sketch both graphs, i.e: y=sin^-1(u) and u=f(x).
After, find values for u in both graphs and find a common restriction. apply this restricition on both graphs to determine D: and R:

Makes it way easier to visualise^

 

[iheartcookies1012]

thanks!

 

[Average Boreduser]

 

[Average Boreduser]

How I'd structure it^

Edit: I just realised I made a mistake in the 2nd qn u asked - I'm guessing u could probs fix that tho.

 

[iheartcookies1012]

omg thank u so much!!

 

[synthesisFR]

View attachment 38857

true composite functions is the best way to set these up

btw ur graph for the second one is incorrect tho

 

[Average Boreduser]

true composite functions is the best way to set these up

btw ur graph for the second one is incorrect tho

wait hold up where'd I go wrong>?

I initially thought I did smthn wrong in the restriction u>=0 for the u=x^2 graph but realised its on y-axis so it can't be all real

 

[synthesisFR]

wait hold up where'd I go wrong>?

the domain
x^2 is positive but that doesn't mean x has to be
negative x values just become positive so the graph exists for [-1,0] as well
it would be a horizontally mirrored version of the part on the right side of the graph

 

[Average Boreduser]

the domain
x^2 is positive but that doesn't mean x has to be
negative x values just become positive so the graph exists for [-1,0] as well
it would be a horizontally mirrored version of the part on the right side of the graph

Oh nvm I understand now lmfao. (mbmb)