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I came across a multiple choice problem where a function $f(x) = \frac{x^2 - 1}{x+1} - x$ is given. One has to choose the statement that is correct about the function. The different statements about the function included:

  • (1) the function increases in proportion to $x^2$.
  • (2) the function increases in proportion to $x$
  • (3) the function is constant.

Obviously the function is constant, but my questions is what the "increases in proportion to" means. I haven't come across it before, and was wondering if it is standard for something. I would like so see an example of a function that can be said to satisfy this statement.

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Please give the whole problem, or we don't know the entire context. Meaning always depends on context. –  Thomas Andrews Jul 6 '12 at 18:14
    
@ThomasAndrews: ok will do –  Thomas Jul 6 '12 at 18:14
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3 Answers

up vote 2 down vote accepted

To say "$f$ increases in proportion to $g$" is just another way of saying "$f$ is proportional to $g$", i.e., $f$ is equal to $g$ times a nonzero constant.

(In other contexts this may only be meant approximately, but the "literal" usage is that it is equal to $g$ times a constant.)

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Would the non-zero constant have to be positive since it is increases ? –  Thomas Jul 6 '12 at 18:25
    
Probably, yes. On the other hand, if $f=-2g$, you probably wouldn't say that "f decreases in proportion to g", because that would probably be taken to mean that f is inversely proportional to g. And it would still be increasing in absolute value. So, probably, but depends on context. –  Harry Altman Jul 7 '12 at 20:07
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A "strict" meaning of:

$f(x)$ increases in proportion to $g(x)$

would be $$f(x)=a g(x)$$ for some constant $a$ (presumably positive.)

There are plenty of looser meanings, but that would depend on a specific definition. For example, if $\lim_{x\to\infty} f(x)/g(x)$ exists, you might say this is true.

If the problem comes from a book, it is possible that the book defines the terminology in a preceding section. If the problem comes from a teacher's quiz, it might have been something defined in class. I'd stick with either the strict definition or the limit definition if this occurred without context, as it appears to do in this problem.

Where did you encounter this problem? It could just be sloppy language used by a test writer.

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I was helping someone with an ACT test where the question came up. –  Thomas Jul 6 '12 at 18:34
    
An actual ACT test, or a practice test from a book? –  Thomas Andrews Jul 6 '12 at 18:38
    
I am not sure about that –  Thomas Jul 6 '12 at 18:38
    
The reason I ask is that if it was in an actual ACT, it probably was vetted, in which case, there is probably a meaning for this term that is "standard" in high school education guides somewhere. But if it was a practice test, it is more likely sloppy language. –  Thomas Andrews Jul 6 '12 at 19:01
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To answer your question about examples where $f(x)$ would be proportional to $x$, and $x^2$, we only need to slightly modify the original function.

$f(x) = \frac {x^2-1}{x+1} $ is interesting between $x=-2$ and $x=2$, but as $x$ becomes really large or really small, the "-1" term and the "+1" term are insignificant and the expression is approximately $\frac {x^2}{x}$ or $x$. This would be proportional to $x$.

If $f(x) = \frac{x^3-1}{x+1} - x$, then as $x$ becomes really large or really small, the function breaks down to $\frac{x^3}{x} -x $ or $x^2$. This would be proportional to $x^2$ for much of $x$.

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So actually $f(x)$ reduces to $-1$ (factor the numerator). –  Thomas Jul 6 '12 at 22:17
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