Order of permutations in transformations of functions

[opus]

Homework Statement

This is for College Algebra.

Describe what is happening to the graph of the function $f\left(x\right)=\sqrt {1 - x}+2$.

Homework Equations

N/A

The Attempt at a Solution

This can be rewritten as $f\left(x\right)=\sqrt {-x + 1}+2$
My conundrum:
Both my text, as well as my instructor, when going into detail about the order of permutations of a function transformation, have said to "start from the inside, and work out".
Both the $-x$ as well as the $+1$ are "inside" the core operator. So which takes precedence, the $-x$ with a reflection across the y-axis, or the $+1$ with a horizontal shift of one unit to the left?
I've tried both scenarios, and they of course give different results, each as a reflection across the y-axis of the other.

 

[mfb]

Describe what is happening to the graph of the function $f\left(x\right)=\sqrt {1 - x}+2$

What do you mean by "happening"? It is just a graph.

Both my text, as well as my instructor, when going into detail about the order of permutations of a function transformation, have said to "start from the inside, and work out".
Both the $-x$ as well as the $+1$ are "inside" the core operator. So which takes precedence, the $-x$ with a reflection across the y-axis, or the $+1$ with a horizontal shift of one unit to the left?

You want to recreate the graph by shifting/mirroring the graph of $g(x)=\sqrt{x}$?
It does not matter, as long as you keep it consistent.
You can first mirror it ($x \to -x$) and then shift it by 1 to the right ($x \to x-1$ and therefore $-x \to -(x-1) = -x+1$).
You can first shift it by 1 to the left ($x \to x+1$) and then mirror it ($x \to -x$ and therefore $x+1 \to -x+1$).

 

[Stephen Tashi]

the order of permutations of a function transformation

That is unusual terminology. Is it actually used in your textbooks? Can you give a link or reference where the terminology is defined?

 

[epenguin]

Rather than fitting this into some mould, which I somehow doubt you are really required to, I would just answer the question.
For what range of x does this f(x) have real values?
After which it may be helpful for visualisation to switch the variables x and f(x) (which you may think of as ‘y’) around (just don’t let any formulae fool you into forgetting the first bit).

 

[SammyS]

I suspect that OP has presented this exercise in his own words.

Here's a snip of an example exercise:

from "College Algebra", 9th Ed. by Ron Larson, published by Cengage.

 

[SammyS]

Homework Statement

This is for College Algebra.

Describe what is happening to the graph of the function $f\left(x\right)=\sqrt {1 - x}+2$.

A better statement of this problem may be:
Consider the basic function, $\ g(x)=\sqrt{x\,}\, .$
What sequence of transformations on the graph of $\ y=g(x)\$ will result in the graph of $\ y=f(x)\,,\$ where $\ f\left(x\right)=\sqrt {1 - x}+2\,?$

Homework Equations

N/A

The Attempt at a Solution

This can be rewritten as $f\left(x\right)=\sqrt {-x + 1}+2$
My conundrum:
Both my text, as well as my instructor, when going into detail about the order of permutations of a function transformation, have said to "start from the inside, and work out".
Both the $-x$ as well as the $+1$ are "inside" the core operator. So which takes precedence, the $-x$ with a reflection across the y-axis, or the $+1$ with a horizontal shift of one unit to the left?
I've tried both scenarios, and they of course give different results, each as a reflection across the y-axis of the other.

I assume that by saying "start from the inside, and work out", they mean that you should follow "order of operations" strictly, as if you were to evaluate the function for a particular numerical value of $\ x\,. \$

This implies that starting with $\ x\,, \$ you take its opposite, which is $\ -(x)\,. \$

If you then add 1 to $\ x\,.$ that gives $\ -(x+1)\,, \$ which is $\ -x-1\,, \$ not $\ -x+1\,. \$

So for this order of transformations, you should subtract 1 from $\ x \$ itself, not from $\ -x\,. \$ That is a shift right rather than left.
All is well.... or is it? Why does the other order also work out? → Add 1, then switch the sign only on $\ x\,. \$

Well, if you have the function: $\ h(x)=\sqrt{x+1\,}\,,\$ then $\ h(-x)=\sqrt{(-x)+1\,}\,.\$ You know this gives a reflection across the $\ y$-axis

 

[Mark44]

order of permutations of a function transformation

As already noted, these aren't usually called "permutations," but rather, transformations.

The transformations fall into three categories: compressions/expansions, reflections, and translations (or shifts).Starting from a basic, untransformed function, it makes a difference in the order of "building in" transformations to get to the function you're interested in. The order that works is to add in compressions/expansions, then reflections, and finally translations.
For example, let's look at $y = g(x) = \sqrt{2x - 3}$. Here the basic, untransformed function is $y = f(x) = \sqrt x$.
If we decide that the graph of g is the translation to the right of the graph of $y = \sqrt{2x}$ by three units, we'll get an incorrect graph. The point $(1, \sqrt 2)$ is on the graph of $y = \sqrt{2x}$, so shifting it to the right by 3 should give us the point $(4, \sqrt 2)$. Instead, g(4) is $\sqrt 5$, so this analysis is obviously flawed.

The proper way is to look at the compression first, then the translation. IOW, look at $g(x) = \sqrt{2(x - 3/2)}$. Starting from $y = f(x) = \sqrt x$, we build in the compression as $y = f(2x) = \sqrt{2x}$. This graph goes through the point $(1, \sqrt 2)$. Next we build in the translation: $y = f(2(x - 3/2)) = \sqrt{2(x - 3/2)}$. Translating the point $(1, \sqrt 2)$ to the right by 3/2 units gives is the point $(5/2, \sqrt 2)$. You can easily verify that $\sqrt{2(5/2 - 3/2)} = \sqrt 2$.

 

[epenguin]

Ah so this sort of thing is what the question meant?
Looks like I misunderstood what the question was and would have got 0 marks, though I know what the function looks like.
How boring!

 

[opus]

Thanks for all the responses!

I could've worded it better. This wasn't directly out of a text, but rather a provided list of exercises to use to take notes on the actual lecture. It's a little difficult to understand what he's saying, so the word he used could possibly not have been "permutations" but something sounding similar. To try to put it more simply, if you have a basic function, such as $f\left(x\right)=\sqrt {x}$, what must be done to that basic function, graphically, to get the provided equation. For example, shift right three units, reflect across the x axis, stretch the graph by a factor of 2, etc...

What do you mean by "happening"? It is just a graph.
You want to recreate the graph by shifting/mirroring the graph of $g(x)=\sqrt{x}$?
It does not matter, as long as you keep it consistent.
You can first mirror it ($x \to -x$) and then shift it by 1 to the right ($x \to x-1$ and therefore $-x \to -(x-1) = -x+1$).
You can first shift it by 1 to the left ($x \to x+1$) and then mirror it ($x \to -x$ and therefore $x+1 \to -x+1$).

Are you saying that the order underneath the radical does not matter? When I graph them separately, they look entirely different as a final result when I do the horizontal shift first or the reflection first. Am I misusing the negative symbol?

Rather than fitting this into some mould, which I somehow doubt you are really required to, I would just answer the question.
For what range of x does this f(x) have real values?
After which it may be helpful for visualisation to switch the variables x and f(x) (which you may think of as ‘y’) around (just don’t let any formulae fool you into forgetting the first bit).

I understand the range part. That is, the graph is a visual representation of the values $f\left(x\right)$ in relation to $x$. From the graph, I could see what the domain and range of the function is. For example, with $f\left(x\right)=\sqrt {x}$, I can see that the domain of the function is $\left[0,∞\right)$ and the range is the same. But the actual actions the the basic function has to undergo to get to the provided function in the OP is what I believe the question is asking me.

I suspect that OP has presented this exercise in his own words.
The problem is from a "guided notes" worksheet, in which it has problems that the students take notes on, while the instructor goes over the problem.
Here's a snip of an example exercise:

from "College Algebra", 9th Ed. by Ron Larson, published by Cengage.

A better statement of this problem may be:
Consider the basic function, $\ g(x)=\sqrt{x\,}\, .$
What sequence of transformations on the graph of $\ y=g(x)\$ will result in the graph of $\ y=f(x)\,,\$ where $\ f\left(x\right)=\sqrt {1 - x}+2\,?$

I assume that by saying "start from the inside, and work out", they mean that you should follow "order of operations" strictly, as if you were to evaluate the function for a particular numerical value of $\ x\,. \$

This implies that starting with $\ x\,, \$ you take its opposite, which is $\ -(x)\,. \$

If you then add 1 to $\ x\,.$ that gives $\ -(x+1)\,, \$ which is $\ -x-1\,, \$ not $\ -x+1\,. \$

So for this order of transformations, you should subtract 1 from $\ x \$ itself, not from $\ -x\,. \$ That is a shift right rather than left.
All is well.... or is it? Why does the other order also work out? → Add 1, then switch the sign only on $\ x\,. \$

Well, if you have the function: $\ h(x)=\sqrt{x+1\,}\,,\$ then $\ h(-x)=\sqrt{(-x)+1\,}\,.\$ You know this gives a reflection across the $\ y$-axis

This may be where I made my error, and your wording of the problem was much better and made more sense.

As already noted, these aren't usually called "permutations," but rather, transformations.

The transformations fall into three categories: compressions/expansions, reflections, and translations (or shifts).Starting from a basic, untransformed function, it makes a difference in the order of "building in" transformations to get to the function you're interested in. The order that works is to add in compressions/expansions, then reflections, and finally translations.
For example, let's look at $y = g(x) = \sqrt{2x - 3}$. Here the basic, untransformed function is $y = f(x) = \sqrt x$.
If we decide that the graph of g is the translation to the right of the graph of $y = \sqrt{2x}$ by three units, we'll get an incorrect graph. The point $(1, \sqrt 2)$ is on the graph of $y = \sqrt{2x}$, so shifting it to the right by 3 should give us the point $(4, \sqrt 2)$. Instead, g(4) is $\sqrt 5$, so this analysis is obviously flawed.

The proper way is to look at the compression first, then the translation. IOW, look at $g(x) = \sqrt{2(x - 3/2)}$. Starting from $y = f(x) = \sqrt x$, we build in the compression as $y = f(2x) = \sqrt{2x}$. This graph goes through the point $(1, \sqrt 2)$. Next we build in the translation: $y = f(2(x - 3/2)) = \sqrt{2(x - 3/2)}$. Translating the point $(1, \sqrt 2)$ to the right by 3/2 units gives is the point $(5/2, \sqrt 2)$. You can easily verify that $\sqrt{2(5/2 - 3/2)} = \sqrt 2$.

So this is the order that we have to "sketch" the graph when we start from it's basic form? An order of operations of sorts?

 

[opus]

Ah so this sort of thing is what the question meant?
Looks like I misunderstood what the question was and would have got 0 marks, though I know what the function looks like.
How boring!

You would've gotten it right. My writing the question was the problem as I had to kind of put it into my own words as it wasn't really an "official" homework problem but rather a notes handout.

 

[Mark44]

So this is the order that we have to "sketch" the graph when we start from it's basic form? An order of operations of sorts?

Yes.
Starting with a basic function, build in compressions/expansions (if any), then reflections (if any), and finally the translations (if any).It's always a good idea to see what happens to a specific point on the graph of the basic function as it is transformed by the various transformations. I did that in my previous post, following the point (1, 1) on the graph of $y = \sqrt x$ through the compression on $y = \sqrt{2x}$ with $(1, \sqrt 2)$, and finally to $y = \sqrt{2(x - 3/2)}$, with $(5/2, \sqrt 2)$.

 

[mfb]

Are you saying that the order underneath the radical does not matter? When I graph them separately, they look entirely different as a final result when I do the horizontal shift first or the reflection first. Am I misusing the negative symbol?

I directly showed the two ways to get the same result, and if you check them they will lead to the same answer. What is unclear?