Graphs of Inverse Circular Functions

[nayano11]

Question: Given that [math]f(x)=\sin(2x)+1[/math], determine the equation of [math]f^{-1}(x)[/math] and hence sketch both functions.

Attempt of response: To obtain the inverse function, I interchange x and y and then make y the subject, yielding:
[math]f^{-1}(x)=\frac{1}{2}\arcsin( x-1)[/math]
When graphing [math]f(x)[/math] and [math]f^{-1}(x)[/math] on my calculator, I get the following: (where red is [math]f(x)[/math] and [math]f^{-1}(x)[/math])

However, my teacher sketches it differently:


Is my teacher correct? For what reason would my teacher have to change the domain? Is it because the functions must intersect?

Apologies if I've written this weirdly. This is my first post.

 

[topsquark]

Question: Given that [math]f(x)=\sin(2x)+1[/math], determine the equation of [math]f^{-1}(x)[/math] and hence sketch both functions.

Attempt of response: To obtain the inverse function, I interchange x and y and then make y the subject, yielding:
[math]f^{-1}(x)=\frac{1}{2}\arcsin( x-1)[/math]
When graphing [math]f(x)[/math] and [math]f^{-1}(x)[/math] on my calculator, I get the following: (where red is [math]f(x)[/math] and [math]f^{-1}(x)[/math])
View attachment 33318
However, my teacher sketches it differently:
View attachment 33319

Is my teacher correct? For what reason would my teacher have to change the domain? Is it because the functions must intersect?

Apologies if I've written this weirdly. This is my first post.

I agree with your graph. I have no idea what your instructor graphed.

-Dan

 

[nayano11]

I agree with your graph. I have no idea what your instructor graphed.

-Dan

I believe she was under the impression that the graph had to have a point of intersection. I suppose she was wrong. Thank you for the reassurance.

 

[Dr.Peterson]

Question: Given that [math]f(x)=\sin(2x)+1[/math], determine the equation of [math]f^{-1}(x)[/math] and hence sketch both functions.

Attempt of response: To obtain the inverse function, I interchange x and y and then make y the subject, yielding:
[math]f^{-1}(x)=\frac{1}{2}\arcsin( x-1)[/math]
When graphing [math]f(x)[/math] and [math]f^{-1}(x)[/math] on my calculator, I get the following: (where red is [math]f(x)[/math] and [math]f^{-1}(x)[/math])
View attachment 33318
However, my teacher sketches it differently:
View attachment 33319

Is my teacher correct? For what reason would my teacher have to change the domain? Is it because the functions must intersect?

Apologies if I've written this weirdly. This is my first post.

The problem is that the problem doesn't state a domain! Did you omit something?

The function f as stated has all real numbers as its domain, and is not invertible! In order to invert it, you have to restrict the domain. You and your teacher chose different restricted domains; you probably chose yours without being aware of it, by blindly using the arcsine and assuming the range of your inverse is the domain of the given function.

What did your teacher give as the equation of the inverse function? And was any further explanation given for that choice?

 

[topsquark]

The function f as stated has all real numbers as its domain, and is not invertible! In order to invert it, you have to restrict the domain.

Ooooooh! I get the graph now!

-Dan

 

[Steven G]

I agree 100% with Dr Peterson. There needed to be a domain given. I didn't look too closely but possibly your teachers graphs are correct but with a different domain from yours.

Having said that your graph is correct for YOUR domain.
I am concerned that you said For what reason would my teacher have to change the domain? I just have one question for you. Which domain did she change from? There are infinitely many Full domains where the original function has an inverse

Another concern I have is that you chose a domain. You can't do that! The author of the problem needs tell you the domain, not you.

I personally would have said that the original function was not invertible and be done. I have done this a few times on exams and had always gotten full credit because I was correct.

 

[nayano11]

The problem is that the problem doesn't state a domain! Did you omit something?

The function f as stated has all real numbers as its domain, and is not invertible! In order to invert it, you have to restrict the domain. You and your teacher chose different restricted domains; you probably chose yours without being aware of it, by blindly using the arcsine and assuming the range of your inverse is the domain of the given function.

What did your teacher give as the equation of the inverse function? And was any further explanation given for that choice?

The problem I posted is exactly like that my teacher had given me. I did not omit anything. I believe we were meant to use arcsine because the equation I obtained for the inverse function is the same my teacher had gotten, so she assumed it'd be arcsine. We were taught that the range of our inverse is the domain of the given function and that the range of arcsine is [math]-\frac{\pi}{2}\le x \le \frac{\pi}{2}[/math].

 

[nayano11]

The problem I posted is exactly like that my teacher had given me. I did not omit anything. I believe we were meant to use arcsine because the equation I obtained for the inverse function is the same my teacher had gotten, so she assumed it'd be arcsine. We were taught that the range of our inverse is the domain of the given function and that the range of arcsine is [math]-\frac{\pi}{2}\le x \le \frac{\pi}{2}[/math].

Apologies I'd written the range incorrectly.

 

[Dr.Peterson]

Question: Given that [math]f(x)=\sin(2x)+1[/math], determine the equation of [math]f^{-1}(x)[/math] and hence sketch both functions.

Attempt of response: To obtain the inverse function, I interchange x and y and then make y the subject, yielding:
[math]f^{-1}(x)=\frac{1}{2}\arcsin( x-1)[/math]

The problem I posted is exactly like that my teacher had given me. I did not omit anything. I believe we were meant to use arcsine because the equation I obtained for the inverse function is the same my teacher had gotten, so she assumed it'd be arcsine. We were taught that the range of our inverse is the domain of the given function and that the range of arcsine is [math]-\frac{\pi}{2}\le x \le \frac{\pi}{2}[/math].

Are you saying the teacher gave the same equation you show as the answer, but graphed it differently? That makes no sense, unless perhaps the graph was copied from an answer book, and was for the problem including a specified domain, which was omitted in the statement of the problem. As we've said, the problem as stated is wrong.

If that graph is the correct answer, then the problem appears to be this:

Given that [imath]f(x)=\sin(2x)+1,\frac{\pi}{4}\le x\le\frac{3\pi}{4}[/imath], determine the equation of [imath]f^{-1}(x)[/imath] and hence sketch both functions.​

Then the solution would be

[imath]f^{-1}(x)=\frac{\pi}{2}-\frac{1}{2}\arcsin( x-1)[/imath]​

Here is the graph of these two equations:

​

 

[nayano11]

Are you saying the teacher gave the same equation you show as the answer, but graphed it differently? That makes no sense, unless perhaps the graph was copied from an answer book, and was for the problem including a specified domain, which was omitted in the statement of the problem. As we've said, the problem as stated is wrong.

If that graph is the correct answer, then the problem appears to be this:

Given that [imath]f(x)=\sin(2x)+1,\frac{\pi}{4}\le x\le\frac{3\pi}{4}[/imath], determine the equation of [imath]f^{-1}(x)[/imath] and hence sketch both functions.​

Then the solution would be

[imath]f^{-1}(x)=\frac{\pi}{2}-\frac{1}{2}\arcsin( x-1)[/imath]​

Here is the graph of these two equations:

View attachment 33325​

You're right. Thank you for the help. One more question: other than graphically, how did you get the equation for the inverse function? I see that the point of inflection is translated 1 unit to the right and pi/2 units up though not how it was reflected. Are there any algebraic means?

 

[Dr.Peterson]

One more question: other than graphically, how did you get the equation for the inverse function? I see that the point of inflection is translated 1 unit to the right and pi/2 units up though not how it was reflected. Are there any algebraic means?

Ah! You took the bait. (I chose not to explain how I got that, and see if you asked.)

And looking at my notes, I see that I actually got it wrong initially and corrected it when I made the graph! That is, I didn't try too hard to do it algebraically; in the end I just knew I had to select the appropriate part of the inverse graph, like this:

​

What I generally do algebraically is to include the restricted domain in my work, changing

[imath]y=\sin(2x)+1,\frac{\pi}{4}\le x\le\frac{3\pi}{4}[/imath]​

to

[imath]x=\sin(2y)+1,\frac{\pi}{4}\le y\le\frac{3\pi}{4}[/imath]​

then solving for y with that restricted range in mind. In this case, I would initially find the principal solution,

[imath]\displaystyle y=\frac{\arcsin( x-1)}{2}[/imath]​

and then think about how to modify that. One approach here is to think about how the arcsin has to be transformed for quadrants 2 and 3: The angle is the supplement of the principal angle. So I'd write

[imath]\displaystyle y=\frac{\pi-\arcsin( x-1)}{2}[/imath]​

But to be honest, after doing that I'd want to check carefully to make sure it works!

 

[nayano11]

Ah! You took the bait. (I chose not to explain how I got that, and see if you asked.)

And looking at my notes, I see that I actually got it wrong initially and corrected it when I made the graph! That is, I didn't try too hard to do it algebraically; in the end I just knew I had to select the appropriate part of the inverse graph, like this:

View attachment 33327​

What I generally do algebraically is to include the restricted domain in my work, changing

[imath]y=\sin(2x)+1,\frac{\pi}{4}\le x\le\frac{3\pi}{4}[/imath]​

to

[imath]x=\sin(2y)+1,\frac{\pi}{4}\le y\le\frac{3\pi}{4}[/imath]​

then solving for y with that restricted range in mind. In this case, I would initially find the principal solution,

[imath]\displaystyle y=\frac{\arcsin( x-1)}{2}[/imath]​

and then think about how to modify that. One approach here is to think about how the arcsin has to be transformed for quadrants 2 and 3: The angle is the supplement of the principal angle. So I'd write

[imath]\displaystyle y=\frac{\pi-\arcsin( x-1)}{2}[/imath]​

But to be honest, after doing that I'd want to check carefully to make sure it works!

Great! Thanks so much for the help. I now know what to do when approaching these problems.