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If $f(x) = \frac{x^2 − 4}{x − 2}$ and $g(x) = x + 2$, then we can say the functions $f = g$

I think this is false but I cannot prove it.

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What is $f(2)$? – user61527 May 9 '14 at 1:02
yes, you are right. – user148834 May 9 '14 at 1:25
up vote 1 down vote accepted

Some teachers might tell you these functions are the same. Other teachers will tell you that they are not. Both viewpoints have merit. However, usually those espousing equality are just being sloppy.

$$f(x) = \dfrac{x^2-4}{x-2} = \dfrac{(x-2)(x+2)}{x-2} = x+2=g(x)$$

The above statement is true everywhere it is defined (i.e. as long as $x \not= 2$). In a loose sense $f$ and $g$ are the "same".

However, strictly speaking the function $f$ and the function $g$ are equal on the domain of $f$. Yet they are different functions since $g$ is defined at $x=2$ while $f$ is not (they have different domains).

Now for many purposes these two functions can be treated as equal (which is why some textbooks/teachers will say they are the "same"). For example:

$$ \lim\limits_{x \to 2} f(x) = \lim\limits_{x \to 2} \dfrac{x^2-4}{x-2} = \lim\limits_{x \to 2} \dfrac{(x-2)(x+2)}{x-2} = \lim\limits_{x \to 2} x+2 = \lim\limits_{x \to 2} g(x) = g(2) = 2+2=4$$

The reason we can do this with the limit is that the limit concerns points "close to" $x=2$. However, $x$ is never allowed to be exactly equal to $2$ (in the limit). Since we aren't actually working at $x=2$, $f(x)=g(x)$ everywhere we are limiting. :)

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What merits for the "these are the same" viewpoint? – Did Mar 17 '15 at 21:23
@Did It really depends on how you are using functions. For example, if you only care about definite integrals, then single points don't impact results. In such a context, $f$ and $g$ are equivalent. You may as well treat them as equal. Also, pedagogically, there are times that belaboring the point that "$f$ and $g$ are equal expect at $x=2$" will lose & confuse a large number of students. That detail can be (temporarily) swept under the rug in order to draw attention so some other important concept. – Bill Cook Mar 18 '15 at 14:22

$f(2)$ is undefined and $g(2)=4$, so they are different functions.

However for $x\neq2$, $\frac{x^2-4}{x-2}=\frac{(x-2)(x+2)}{x-2}=x+2$. So on the restricted domain $(-\infty,2)\cup(2,\infty)$, they are the same function.

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The functions are not equal because they have different domains.

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In this case you can replace $x$ with a known value in the domain of both functions. Remember you're testing for $f(x)=g(x)$ so for the statement to be true the result after solving for one value of $f$ (say $f(2)$) is going to have to be the same for $g$. To make it clear:

IF $f(x) =g(x)$ is true then:

$f(1) = g(1)$

$f(2) = g(2)$


Hope it helps.

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