Tracking a point, with inverse functions

M

mynamesmurph

Junior Member
Joined
Aug 10, 2014
Messages
51
Suppose point (a,b) is on the graph of a function. Given a defined function, what will the points be? (I'm supposed to match from a given list)

y=f^-1(x-1)
It seems the graph is translated to the right one, and then the inverse function switches the points.

(a,b) becomes (a+1,b) becomes (b,a+1)


y=f^-1(x)+1


Here, it matters which order things are done. If I translate the graph up one first, apply the inverse function I get (a,b) becomes (a,b+1) becomes (b+1,a)

If I use the inverse function first, then translate, I get (a,b) becomes (b,a) becomes (b,a+1).




I was used to doing translations and reflections without inverse functions and I seemed work "inside out" or "outside in"

For example.

y=-f(x)+1

I could either shift up first, then reflect across the x axis, or reflect across the x axis then shift up one.

Any help?
 
mynamesmurph said:
Suppose point (a,b) is on the graph of a function. Given a defined function, what will the points be? (I'm supposed to match from a given list)

y=f^-1(x-1) this is the inverse of f(x) translated one to the right
It seems the graph is translated to the right one, and then the inverse function switches the points. No, 1st inverse then translated. So (a,b) becomes (b,a) becomes (b+1, a)

(a,b) becomes (a+1,b) becomes (b,a+1)


y=f^-1(x)+1


Here, it matters which order things are done. If I translate the graph up one first, apply the inverse function I get (a,b) becomes (a,b+1) becomes (b+1,a). No, you are adding one to f^-1(x). So first get the inverse and then add 1 to the y value of this inverse. So (a,b) becomes (b,a) becomes (b, a+1)
That is: assume f(a) = b. (I will use f" as the symbol for inverse). Then f"(b) = a and f"(b) + 1 = a+1

If I use the inverse function first, then translate, I get (a,b) becomes (b,a) becomes (b,a+1).




I was used to doing translations and reflections without inverse functions and I seemed work "inside out" or "outside in"

For example.

y=-f(x)+1

I could either shift up first, then reflect across the x axis, or reflect across the x axis then shift up one. But this gives different results. Suppose for example that f(2) = 5. If you shift up first, then reflect across the x axis then you get after the shift up one (2,6), After you reflect across the x axis you get (2,-6). On the other hand if reflect across the x axis then shift up one you will get(2,-6) after you reflect (2,6) across the x-axis. Now after you shift up one you get (2,-5). The problem is that you get different points. The order does matter whether you are translating and/or shifting f(x) or f"(x).

Any help?
Click to expand...
You need to do this in a way that is understandable which I translate into you need to use the definition.
Using your example above: y=f"(x)+1. So you need f"(x) and then add 1 to it. In particular you need f"(x)
Assume f(a) = f(b). Then f"(b) = a. Now we add 1 to both sides and get f"(b) + 1 = a+1. So (b, a+b)

Consider this problem: f"(x+2). Assume f(a) = b. Then f"(b) = a. Then f"(b) = f"( [b-2]+2) = a. So (a,b) becomes (b-2,a).
I hope this helps.
 
Last edited:
Jomo said:
You need to do this in a way that is understandable which I translate into you need to use the definition.
Using your example above: y=f"(x)+1. So you need f"(x) and then add 1 to it. In particular you need f"(x)
Assume f(a) = f(b). Then f"(b) = a. Now we add 1 to both sides and get f"(b) + 1 = a+1. So (b, a+b)

Consider this problem: f"(x+2). Assume f(a) = b. Then f"(b) = a. Then f"(b) = f"( [b-2]+2) = a. So (a,b) becomes (b-2,a).
I hope this helps.
Click to expand...


Yes, it did help. I think I have an intuitive understanding of what is going on. I understand why the order of (at least some of the time matters), and more importantly I'm confident of when I need to do it.

However, I'm struggling to understand your further explanations here. The first example makes sense, I think, but the second example I seem to be getting lost.
 
mynamesmurph said:
Yes, it did help. I think I have an intuitive understanding of what is going on. I understand why the order of (at least some of the time matters), and more importantly I'm confident of when I need to do it.

However, I'm struggling to understand your further explanations here. The first example makes sense, I think, but the second example I seem to be getting lost.
Click to expand...
I want to find f"(b) but must write it in the form f"(x+2) ie I want x+2=b. So x=b-2. I guess that is the part that confused you? If not, then please let me know where I lost you
 
mynamesmurph said:
Could you do y=f(-x)
Click to expand...
Yes. But you are the one who needs to learn this, so please reply with your thoughts and your, especially as relates to the answers and explanations you've already been given, so that we can see where you're having trouble. Please be complete. Thank you! ;)
 
I should say I'm only asking so I can better see the pattern. I suppose my problem is actually, What should I be asking? It's hard for me to ask the right questions, when I'm having trouble interpreting the help I'm (generously) being given. Just so you know, I'm not trying to get the answers to my homework, I'm really trying to understand the material. We just started functions, so my knowledge is very shaky. That said, forgive any rambling, etc.

I understand that y=f(-x) is there reflection of the function over y axis. Thus point (a,b) would become (-a,b)

More generally speaking, I understand that when I reflect the graph of a function over the y axis, the order of translations of the x value are important, as shifting first then reflecting will not yield the same results as reflecting then shifting. Additionally, reflections over the x axis and vertical shifts, the order is important.

When I view a translation of some function like this.

y=-f(-x+1)+1

This is my thought process. I see f(x), reflected over the y axis f(-x) shifted to the right f(-x+1) [alternatively, I suppose I could take f(x) shift to the right f(x-1) and then reflect over the y axis like this f(-(x-1)) ] then reflected over the x axis y=-f(-x+1) and finally vertically shift up one -f(-x+1)+1

With regard to inverse functions, point (a,b) must always become (b,a) first since, that is the point we are going to be translating. From there it the same before.

When I see something like

f"(x-1) points (a,b)

I think, the inverse function of x, (b,a) shifted to the right (b+1,a)

f"(-x)+1

The inverse function of x (b,a) reflected over the y axis (-b,a) and then shift up one, (-b,a+1)

-f"(-x)+1

The inverse function of x (b,a) reflected over the y axis (-b,a) reflected over the a axis (-b,-a) shift up one (-b,1-a)




But I haven't seen it explained in the context of assuming that f(a)=b and am trying see the value in doing so.



You need to do this in a way that is understandable which I translate into you need to use the definition.
Using your example above: y=f"(x)+1. So you need f"(x) and then add 1 to it. In particular you need f"(x)
Assume f(a) = f(b). Then f"(b) = a. Now we add 1 to both sides and get f"(b) + 1 = a+1. So (b, a+b)

Consider this problem: f"(x+2). Assume f(a) = b. Then f"(b) = a. Then f"(b) = f"( [b-2]+2) = a. So (a,b) becomes (b-2,a).
I hope this helps.


So, I will try go through this myself and perhaps it reveal any misunderstandings.

You need to do this in a way that is understandable which I translate into you need to use the definition.

I think what is being said is that I need to understand why I'm using the rules that I'm using? I'm not quite sure.

Using your example above: y=f"(x)+1. So you need f"(x) and then add 1 to it. In particular you need f"(x)

Here, we were taking the untranslated inverse function and adding 1

Assume f(a) = f(b). Then f"(b) = a.

So, I think the point here is to use points (a,b) for my x and y values, or input and outputs.

Assume f(a) = f(b). I am a little confused here. I know f"(f(a)) = a
I know on the next example, the assumption is f(a)=b, so is that what was meant on the first instruction? I will assume it is, because then taking the inverse function of both sides, I can see how f"(b) = a

Now we add 1 to both sides and get f"(b) + 1 = a+1. So (b, a+b)

adding 1 to both sides we get f"(b)+1 = a+1
So, our point becomes (b, a+1) (or is it supposed to (b, a+b)? as in the example?)


So, if my assumptions are correct, is the point of all of this, to manipulate the function with (a,b) to make it look like the original to reveal what happens to our points on the graph. In this case, we assume that f(a) = b, So if we manipulate this in a way that makes it identical in the first function, it will show the order in which we should be applying our inverses, reflections and translations. Right? If function has an inverse, (and we can assume it does, because the example is an inverse function), we know f(a)=b will become a=f"(b), reversing our points (a,b) to (b,a). Then we add one to both sides, my input doesn't change, but my output increases by 1. This gives me (b,a+1)




I will continue on with the second example, after I clear up an misunderstandings here.




Also, perhaps you can demonstrate an actual example. I have some disconnect going on here and I'm not sure how to phrase it.

consider the translation to the function

f(x+1)=y

and suppose f(x)=x^2

How can we track point (a,b) here?




After I get this cleared up, I'll work with f(-x)




Sorry and Thanks a million.
 
Last edited:
Are you using f" to indicate the inverse function? That is normally used to indicate the second derivative. It is more common to write the inverse function to f(x) as f^-1(x).
 
You must log in or register to reply here.
Top