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Post by
Interest
Screw math.
Screw academics.
Lol
Post by
Kristopher
Screw math.
Screw academics.
Lol
That too.
Post by
Kristopher
Goddammit, may I have some help with my math homework, again? ...
The functions
f
and
g
are defined by these sets of input and output values.
<order is as it is in the book>
g
= {(1, 2), (-2, 4), (5, 5) (6, -2))
f = {(0, -2), (4, -1) (3, 5), (5, 0)}
a. Find g(f(4)).
b. Find f(g(-2))
c. Find f(g(f(3))).
I believe I know how to solve the problems (substitute for x
), but I'm unsure about my ability to find the equation for the values... I think if I get help with the procedure for one, I can complete the other.
Post by
Heckler
If that's all that's given, it's not enough to find a unique form for f(x) and g(x). There's an arbitrary number of functions that pass through the four provided points (if a "form" were given, then a unique answer might exist, for example, y = Ax+B has a unique solution for A and B given 2 data points. Since you're given 4 data points, you
could
uniquely define a 3rd order polynomial, y = Ax^3+Bx^2+Cx+D -- but this seems beyond the scope of the class, and there's an infinite number of n-th order polynomials for n>3 -- in addition, it would be perfectly legitimate if g(x) was
only
defined on 4 points, and simply undefined elsewhere. This actually seems the most reasonable assumption to make at this point).
I think you might have mis-transposed something in the given information (because part a doesn't work, unless an acceptable answer is "undefined"), but here's how you would do part b.
f(g(-2)): First, find g(-2) -- the ordered pair which begins with -2 provides this answer: (-2,4) means
4
. Now find f(4), which according to the ordered pair (4,-1) is -1. So f(g(-2)) =
-1
.
Post by
Kristopher
I think I wrote them correctly from the book, nothing looks wrong...
I'm not sure if "undefined" is an answer, but I am really starting to think that I'm lacking a lot of knowledge that I should have for this class, or the teacher isn't teaching us the things she's making us do for our homework. . .
Anyway, so if you can get the answer for c. by looking at (-2, 4) than a later problem of "find g(f(2)) where
g = {(1, 2), (-2, 4), (5, 5) (6, -2))
f = {(2, 1), (4, -2), (5, 5), (-2, 6) "
then I would look at (2, 1) and the answer for g(f(2)) would be 1?
Post by
Jubilee
You always just work from the innermost parentheses to the outermost. f(2) = 1, so substitute that in, and g(1) = 2.
So g(f(2) = 2
I'm so bored, let's do something funner than math =P
Post by
Kristopher
I wish I could do that but it's homework, and I haven't been doing so well on my math homework lately... :(
Also, giving up on those other problems for now, "pick your own x-value in the domain of g and find f(g(x))" means pick an x-value that g(x) produces, and substitute it into that function, right?
Nevermind, got that problem out of the way.
Post by
Heckler
I'm not sure what you covered in lecture, so I'll try to make this post a completely self-contained explanation of the topic as I understand it -- read it thoroughly and carefully, I'll do my best to write it well.
So, the ordered pairs mean (input, output). For example, an ordered pair of (3,5) means that f(3) = 5.
The notation:
g = { (1, 2), (-2, 4), (5, 5) (6, -2) }
Means that the function g is
defined
at the points (1, 2), (-2, 4), (5, 5), and (6, -2). It's important to note that without more information, you don't know anything about any other points in g. What I mean by that is, you know g(1), g(-2), g(5), g(6) but you don't know anything about g(50) or g(22). All you can really say is that they're
undefined
.
The next notation to know is:
f(g(x))
Just like other parenthesis, you work from the inside out. This is the same thing as a two part problem: if y = f(g(x)), then you could also say that
z
= g(x), and y = f(
z
). It means you must first find the output value for g, and then plug that output value into f as its input value.
Now, for your problem set, Let's rewrite the "Ordered Pair Definitions" to something more familiar, and then rework the problems.
Instead of writing:
g = { (1, 2), (-2, 4), (5, 5), (6, -2) }
f = { (2, 1), (4, -2), (5, 5), (-2, 6) }
Let's instead write this equivalent definition for f and g:
g(1) = 2; g(-2) = 4; g(5) = 5; g(6) = -2;
f(2) = 1; f(4) = -2; f(5) = 5; f(-2) = 6;
Look over the two boxes above. If you can understand why they're both saying the same thing, then you have a full grasp of this concept I would say.
Now, let's do the problems...
(a) g(f(4))
First (work from the inside out), find f(4). From our box above, f(4) = -2.
Now we can rewrite g(f(4)) as g(-2).
From our box above, we know g(-2) = 4.
So
g(f(4)) = 4
.
(b) f(g(-2))
First (work from the inside out), find g(-2). From our box above, g(-2) = 4.
Now we can rewrite f(g(-2)) as f(4).
From our box above, we know f(4) = -2.
So
f(g(-2)) = -2
.
I think you should be able to get (c) using these tips.
Post by
Jubilee
x is your input not your output. g(x) doesn't produce an x value. The x values in the domain of g are 1, -2, 5, and 6. You pick one of those, solve the function by looking at the corresponding y value, then take that y value and make it the x value of your f function and solve that by look at that number's corresponding y value.
Post by
Interest
Finally.
NEVER HAVE TO DO ANY MORE HEROICS THIS PATCH (on my hunter).
Post by
Kristopher
f(g(f(3))):
f(3) = 5, g(5) = 5, f(5) = 0
is that right?
Post by
Jubilee
3 wasn't in those domains you gave us.
Post by
Heckler
In the first problem it was, I think he's presented two separate definitions for f and g. If the very first f and g definitions you posted are right, then yes, f(g(f(3))) = 0. =)
Post by
Kristopher
3 wasn't in those domains you gave us.
Yes it was? I think?
g = {(1, 2), (-2, 4), (5, 5) (6, -2))
f = {(0, -2), (4, -1) (
3
, 5), (5, 0)}
In the first problem it was, I think he's presented two separate definitions for f and g. If the very first f and g definitions you posted are right, then yes, f(g(f(3))) = 0. =)
Edit: Woohoo, thanks so very much! Now I think I understand THIS set of problems, lol. Woe be me in a few weeks (yay light work due to testing coming up?) when it changes >_>
Post by
Heckler
That also means that the (a) and (b) I worked out in my long post weren't the "right" (a) and (b) though xD
Post by
Jubilee
3 wasn't in those domains you gave us.
Yes it was? I think?
g = {(1, 2), (-2, 4), (5, 5) (6, -2))
f = {(0, -2), (4, -1) (
3
, 5), (5, 0)}
Oh I sawg = {(1, 2), (-2, 4), (5, 5) (6, -2))
f = {(2, 1), (4, -2), (5, 5), (-2, 6)
With those, that is correct.
Post by
Kristopher
That also means that the (a) and (b) I worked out in my long post weren't the "right" (a) and (b) though xD
I didn't write down the answers anyway, I'd rather get a bad grade than pass in work that I didn't do. I followed what you said to do and I got
a. g(f(4))= 2
b. f(g(-2)= 1
and I feel now that those are right.
Post by
Heckler
Ah yes, so that's the one I'm concerned about, (a) g(f(4)) :
f(4) = -1 (
negative
1)
g(-1) is not defined.
I agree with (b) though.
Oops, no I don't. Jubilee is right, i don't quite agree with (b) either (unless there's a copy error in the f and g definitions).
Post by
Jubilee
Based on these numbersg = {(1, 2), (-2, 4), (5, 5) (6, -2))
f = {(0, -2), (4, -1) (3, 5), (5, 0)} Those are not correct
Post by
Kristopher
Oh god damn my eyes. f(4) = 1, not -1. I don't know how I didn't notice that before... at least I'm right now, though! Again, thank you so much. It feels good to actually understand something, after I've been struggling and doing poorly with the class.
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