Help needed with Domian and Range of inverse sine functions (1 Viewer)

[blackops23]

Hi, I have to find the domain and range of g(x)= inv. sin [1/(x^2 + 1)]

Here's what i did:

domain of y=inv.sin(x) is -1<(x)<1
therefore domain of g(x): -1 < [1/(x^2 +1)] < 1

-(x^2 +1) < 1 < (x^2 + 1)

So... what's the correct way of doing it, other than randomly subbing in points?

Thanks guys

 

[jyu]

0<1/(x^2+1)=<1 for all x
so -1=<1/(x^2+1)=<1 for all x

 

[hscishard]

Yea...you're almost there
just solve the 2 inequalities

Oh for the range it's 0 < y <= pi/2 as 1/x^2+1 cannot be 0

 

[blackops23]

Yea...you're almost there
just solve the 2 inequalities

Oh for the range it's 0 < y <= pi/2 as 1/x^2+1 cannot be 0

-x^2 - 1 <= 1
x^2 <= -2
Uh oh...

But when I graph -x^2 -1, it is <= 1 for ALL x, but then with x^2, it is not at all <=-2, so wtf is going on??

What the heck do I with the left end, just ignore it?

x^2 + 1 > 1
x^2 > 0

help please

 

[hscishard]

x^2>=-2
x>= root2 x<=root2
x^2>=0
All real x

It really is just all real x... The graph whould look like a cheap speed bump. It's easy to graph if you do ext2. Just use wolfram

 

[blackops23]

x^2>=-2
x>= root2 x<=root2
x^2>=0
All real x

It really is just all real x... The graph whould look like a cheap speed bump. It's easy to graph if you do ext2. Just use wolfram

... the problem is... I am doing 4u... :S missed out on too much class work so I'm catching up at home... but probably should drop it anyways

Thanks anyway

And also, for x^2 = -2, how can you take the root of a negative? Or do you just solve the graph through graphing rather than algebraically?

 

[hscishard]

... the problem is... I am doing 4u... :S missed out on too much class work so I'm catching up at home... but probably should drop it anyways

Thanks anyway

And also, for x^2 = -2, how can you take the root of a negative? Or do you just solve the graph through graphing rather than algebraically?

You can sketch it easily after you've learnt graphs in 4u.

Oh and I made a mistake for x^2=>-2, it's no solutions. I got so confused from your > and < sign from your working

 

[Bored Of Fail 2]

... the problem is... I am doing 4u... :S missed out on too much class work so I'm catching up at home... but probably should drop it anyways

Thanks anyway

And also, for x^2 = -2, how can you take the root of a negative? Or do you just solve the graph through graphing rather than algebraically?

Check the thread you posted at hscguide

I solved it there algebraically , it is possible to solve it without a graph from wolframa, you obviously wont have access to math software for your hsc exams!

 

[blackops23]

regarding the range of the function, all I did was sub various points in, e.g. x=0, x=1, x=10^20.

However the textbook had a much faster method, but it wasn't all that clear to understand. It went: "range is 0<y<= (pi/2), due to inv.sin(0)=0 and inv.sin(1)=pi/2
I don't really understand where this method orginates from, why does the 0 and 1 determine the range?

Is it due to the range of y=1/(1+x^2)??

 

[Bored Of Fail 2]

regarding the range of the function, all I did was sub various points in, e.g. x=0, x=1, x=10^20.

However the textbook had a much faster method, but it wasn't all that clear to understand. It went: "range is 0<y<= (pi/2), due to inv.sin(0)=0 and inv.sin(1)=pi/2
I don't really understand where this method orginates from, why does the 0 and 1 determine the range?

cnt understand question

 

[blackops23]

cnt understand question

Basically I want to know is how did the textbook calculate the range using inv.sin(0) and inv.sin (1)

 

[Bored Of Fail 2]

What did it say the range was??

I would think it is 0<= y <= pi/2

 

[Bored Of Fail 2]

ok. yes we know that 1/ (1+x^2 ) > 0 for all x

therefore we are putting in only positive values into the inverse sin function ( this is why we restrict it to the only being interested in the interval 0<=x<=1 of the inverse sin graph )

also, remember that the derivative of the inverse sin function is 1/ sqrt ( 1-x^2 ) , which is positive , which means that the graph is monotonically increasing, thus the range can be found by evaluating the function at the endpoints.

similarly , if we had inverse cosine we can use the same method of evaluating the endpoints because the derivative of inverse cosx is -1/sqrt ( 1-x^2) is negative, and montonically decreasing.

 

[blackops23]

What did it say the range was??

I would think it is 0<= y <= pi/2

Yeah is the range, but how one earth can you get the range by just doing: inv.sin(0) = 0 & inv.sin(1) = pi/2. Like wtf, why is it that you can figure out the range of the function through the inverse sine of "0" and "1".

Also another example of this wierd method:

y= inv.sin (x^2 -1)
Quoted from textbook:

"Range: -pi/2 <= y <= pi/2, due to inv.sin(-1) = -pi/2, and inv.sin(1) = pi/2 "

 

[Bored Of Fail 2]

see my above post, it is all about the derivatives and the fact that they are always increasing

thus the least and greatest values are always at the endpoints

 

[cutemouse]

Just use common sense

The domain of 1/(x^2+1) is all real x and the range is 0<y<=1

So the domain of arc sin(1/(x^2+1)) is all real x (since 1/(x^2+1) will always be between -1 and 1 from above)

the range would be 0<y<= pi/2 as arcsin(0)=0 and arcsin(1)=pi/2 (substituting the endpoints of the range of 1/(x^2+1) )

 

[Bored Of Fail 2]

see my above post, it is all about the derivatives and the fact that they are always increasing

thus the least and greatest values are always at the endpoints

is this method clear to you blackops??

it is quite a nice method actually, all I ever used to was remember the domain of the inverse functions because I knew that you can find the range just by subing in the endpoints of the domain into the function

 

[Trebla]

Hi, I have to find the domain and range of g(x)= inv. sin [1/(x^2 + 1)]

Here's what i did:

domain of y=inv.sin(x) is -1<(x)<1
therefore domain of g(x): -1 < [1/(x^2 +1)] < 1

-(x^2 +1) < 1 < (x^2 + 1)

So... what's the correct way of doing it, other than randomly subbing in points?

Thanks guys

For the domain, note that:



Also note that:



This means that using the greatest lower bound and lowest upper bound to find the domain we end up with the statement that



Since this already satisfies the generalised requirement of



then the domain is clearly all real x.

For the range, note that the inverse sine function is always an increasing function. Thus the higher the value of



then the higher the value of



Since



then

 

[cutemouse]

then

Nice explanation Trebla. But I think it's worth noting that your explanation works because the inverse sin function is increasing in its domain.

 

[blackops23]

ok. yes we know that 1/ (1+x^2 ) > 0 for all x

therefore we are putting in only positive values into the inverse sin function ( this is why we restrict it to the only being interested in the interval 0<=x<=1 of the inverse sin graph )

also, remember that the derivative of the inverse sin function is 1/ sqrt ( 1-x^2 ) , which is positive , which means that the graph is monotonically increasing, thus the range can be found by evaluating the function at the endpoints.

similarly , if we had inverse cosine we can use the same method of evaluating the endpoints because the derivative of inverse cosx is -1/sqrt ( 1-x^2) is negative, and montonically decreasing.

So umm... how would that apply to y = inv.sin (cos x)??

And so if the domain is ALL REAL X, you would just use inv.sine(1) and inv.sine(-1) to find the domain??