# Can someone show me how to solve functions that include a set of ordered pairs? Im really stuck on this.

Im unsure of how to go about solving these when given a set of

Given

$f= \{(-2,4),(0,-2),(1,2),(2,3)\}$ and $g(x)= 3x-5$

Find

A) $f(-2)$

B) $(f+g)(2)$

C) $f(g(x))$

D) $f(1)-g(1-k)$

In each case I’m not sure how to calculate the $f$ values.

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What have you tried? What is your definition of a function? –  wj32 Oct 22 '12 at 0:24
First, verify $f$ is a function. Once you have done that you will probably have recalled the way in which one looks up values for a function defined in such a manner, which is all that has to be done. Finally, part C does not make sense to me, perhaps you means g(f(x)). –  peoplepower Oct 22 '12 at 0:25
Hint: $f(2)=3$. –  MJD Oct 22 '12 at 0:58
Part C has its domain restricted to the x values of the pairs given for $f$. –  Ben Oct 22 '12 at 1:11

The part I think you are missing is that $f$ is a function which is defined only at the values -2, 0, 1, and 2, and nowhere else. $f(-2) = 4$, which is the answer to part (A); $f(2) = 3$, from which you can easily solve part (B); $f(1) = 2$, from which you can solve part (D).

This leaves only part (C), to determine $f(g(x))$. $f$ is defined only at -2, 0, 1, and 2, so the composite function $f(g(x))$ is defined only at values of $x$ where $g(x)$ is one of -2, 0, 1, or 2. If $x$ is such that $g(x)$ comes out to something else, say 37, then that is an invalid argument for $f$, and the entire expression $f(g(x))$ is undefined. So what you need to do is make a list of the few values of $x$ that make $g$ yield a good argument for $f$, and then you can describe the behavior of the resulting $f(g(x))$ function in a way similar to the way that $f$ itself was described.

In principle, all functions can be described as a list (possible a very large infinite list) of ordered pairs in this way. In some contexts in mathematics we take an ordinary function like $g(x) = 3x -5$ and define it as the infinite set $$\{\ldots, (-2, -11), (-1, -8), (0, -5), (1, -2), \left(1\frac13, -1\right), \\ (2, 1), (\pi, 3\pi-5), \ldots, (57.89, 168.67), \ldots \}.$$ The function $f$ here is no different in principle; it just has a much smaller domain.

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A) $f(-2) = 4$

B)$(f+g) (2) = 4$

D) $f(1)-g(1-k) = 2 - (3(1-k) - 5)$

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Can you explain part C? –  wj32 Oct 22 '12 at 0:29
Part C) f(g(x)) this is exactly how it is listed on my assignment. This is what confused me the most. :/ –  John E. Oct 22 '12 at 0:55