Inverse functions (1 Viewer)

[micuzzo]

hi
i have a simple question... to find an inverse function of a fuction that is not 1-1 we have to restrict the domain... but if u do find the inverse function and it has a root... how do we knwo whether to limit the domain to be increading or decreasing... eg on fitz pg. 138 question 22... i think they have limited the orginal function to be decreasing... how do we know if it should be decreasing or increasing since in the previous Q itwas increasing


thanks
...hope it makes sense...typing toofast

 

[lolokay]

you limit the domain first off

with the inverse, the range must be the same as the domain you chose (the range and domain swap around, when you interchange the x and y), so choose the root that satisfies this

eg. if you started by limiting the domain to x<0
and for the inverse you got to
y = +-squareroot[something]
then you know that y<0, so choose the negative square root

 

[jet]

well normally you just do an increasing domain. Thats what I always did. I guess you would just look at the function and found the values of the domain which you thought would make it translate into a 'good' inverse function. By good i mean one you can work well with and understand.
For example, if you restricted it to a domain that covered very extreme values of the scale you would have a lot more trouble than a person who would choose a domain near the origin.
All of this of course would rest on the fact that the turning points are in 'good' places as well.

 

[micuzzo]

^^thanks guys... but is there some kind of a set rule i should follow or in my answer should i just say wat i am limiting the domain to??

 

[Aerath]

Limit it so it covers as much as the domain as possible.

However, just by convention, if you have a choice between x<-2 and x>2, choose x>2, purely because it covers the positive x axis. Also, if you have a choice between -1<x<1 and 2<x<4, obviously go with -1<x<1, because it covers x = 0.

 

[micuzzo]

so does it matter if its increasing or decreasing??

 

[youngminii]

so does it matter if its increasing or decreasing??

Not really no, but it's convention(as in, this is the way that's normally used) to use the 'more' positive domain.

 

[Aerath]

so does it matter if its increasing or decreasing??

Not a factor unless the question specifies so (which I've never seen before)

 

[micuzzo]

^^ thanks guys =]

 

[tommykins]

^^ thanks guys =]

why would you ban water

 

[micuzzo]

why would you ban water

i would never ban water... but others may want to ban DHMO after they google it...


ok i have another question...

can someone explain how to solve arcsin (-1/root2)... i dont really get the restriction thing

 

[Timothy.Siu]

i would never ban water... but others may want to ban DHMO after they google it...


ok i have another question...

can someone explain how to solve arcsin (-1/root2)... i dont really get the restriction thing

u use the calculator and it says -45 degrees on mine,

 

[micuzzo]

u use the calculator and it says -45 degrees on mine,

um... are u joking...??

i mean how to show working...u no like with quadrants...

 

[Timothy.Siu]

um... are u joking...??

i mean how to show working...u no like with quadrants...

u can draw a triangle on the Cartesian plane

 

[micuzzo]

thats the part im stuck with... i dont no where to draw it 1st 2n3rd or 4th... text books dont explain clearly

 

[Timothy.Siu]

here

 

[lolokay]

inverse sin is only in the 1st and 4th quadrants

if it's negative it has to be in 4th

 

[micuzzo]

ok i get that... just not wher to put it... can u do tan[arcsin(-root3/2)]

i no i can do with calc... just working confusing

 

[micuzzo]

inverse sin is only in the 1st and 4th quadrants

if it's negative it has to be in 4th

yup ok thats wat im confused with... 1 do we only give it 1 value not 2

 

[lolokay]

you only get one sin^-1x value for every x value

its range is -pi/2 =< sin^-1x =< pi/2