5

Suppose I have a function defined as $$F(x)= \frac{x^3-4x^2+4x}{x^2-4}$$

Now I want to find the value of $F(2)$. I can do it in 2 ways:

  1. Put $x=2$ and solve the function. It will give:

    $$F(2)=\frac{0}{0}$$ which is not defined.

  2. Solve $F(x)$ first and then put $x=2$.

    $$F(x)= \frac{x(x-2)^2}{(x-2)(x+2)}=\frac{x(x-2)}{x+2}$$

    It will give $${F(2)=\frac{0}{4}}$$ which is zero.

How can zero equal not defined?

Vidyanshu Mishra
  • 10,143
  • 1
    In step 2. you canceled out $x-2$ which is $0$ when $x=2$. – dxiv Nov 23 '16 at 20:13
  • 3
    Oops, I got it.Thanks @dxiv, I think I have made similar error which people make while they prove 2=1 – Vidyanshu Mishra Nov 23 '16 at 20:15
  • @dxiv wait, What is the problem in canceling x-2 by x-2. I haven't mentioned that I m going to find F(2) while I cancelled x-2 by x-2?? – Vidyanshu Mishra Nov 23 '16 at 20:19
  • 3
    You cannot cancel $0$. So the moment you did cancel $x-2$, you assumed that $x \ne 2$. The expression you got after canceling is not valid for $x=2$ because you assumed $x \ne 2$ while deriving it. – dxiv Nov 23 '16 at 20:21
  • Got it. Thanks. – Vidyanshu Mishra Nov 23 '16 at 20:21
  • $+1$ for the question. I don't know why someone down-voted it. Please look at my answer below. $\qquad$ – Michael Hardy Nov 23 '16 at 20:38
  • Site is full of those people who make questions neutral (by downvoting it if it doesn't deserve an upvote or by upvoting it if it doesn't deserve downvote. Anyway, this is not the matter. – Vidyanshu Mishra Nov 23 '16 at 20:40
  • I like your attitude, The Lone Wolf. I've only known you for posting substantive questions that clearly show you've put time into it. You'll go far with your attitude: putting priority on learning. Unfortunately, too many askers become discouraged with the responses they receive here. So I'm glad the voting "thang" doesn't matter to you as much as finding the help you need. – amWhy Nov 23 '16 at 20:55
  • One comment: When something is undefined in math, we don't say it's "equal to undefined", we just just say it's "undefined". – MathematicsStudent1122 Nov 23 '16 at 21:56
  • @THELONEWOLF.: Expect to see this kind of thing when you study "limits" in Calculus. See, for instance, this question, and perhaps in particular, my answer (but also others). – Blue Nov 24 '16 at 12:09
  • @blue,I saw this function when I was dealing with limits. Though I am not so good in that topic.:) :) – Vidyanshu Mishra Nov 24 '16 at 13:05

4 Answers4

3

Remember that functions consist not only of rule of assignment but domain and codomain as well. So, let $$f\colon\Bbb R\setminus\{\pm2\}\to\Bbb R,\quad f(x) = \frac{x^3-4x^2+4x}{x^2-4}$$ and $$g\colon\Bbb R\setminus\{-2\}\to\Bbb R,\quad g(x)=\frac{x(x-2)}{x+2}$$

Just by inspecting domains you can immediately see that these are not equal functions. However, what you did show by your simplification is that restriction of $g$ on $\Bbb R\setminus\{\pm 2\}$ is equal to $f$, i.e. $f(x) = g(x)$ for all $x\neq\pm 2$.

This is rather a common mistake that I believe is due to how algebraic expressions are taught in high school, completely ignoring defining context in which this is allowed. It is allowed, for example, when one takes a limit of the functions:

$$\lim_{x\to 2} f(x) = \lim_{x\to 2} g(x) = g(2) = 0$$

But, $\lim_{x\to 2}f(x) = 0$ does not imply that $f(2) = 0$.

Another trivial example could be functions $f(x) = \frac xx$ and $g(x) = 1$ which are equal at all points but $x=0$. Again, natural domains of $f$ and $g$ are not equal.

Ennar
  • 23,082
1

What you have found is a simplification for $F(x)$, provided $x\neq 2, x\neq -2$. The denominator $(x^2 - 4)$ of the original function makes it undefined at $x = 2, \;x=-2:$ $(2^2-4) = (-2)^2 - 4 = 0$

So your simplification

$$F(x)= \frac{x(x-2)^2}{(x-2)(x+2)}=\frac{x(x-2)}{x+2}$$ is valid, $\forall x \in \mathbb R \setminus\{-2, 2\}$.

amWhy
  • 209,954
1

The reason is simply that

$$ \frac{x^3-4x^2+4x}{x^2-4}$$ and $$\frac{x(x-2)}{x+2}$$

are two different expressions.

Their values indeed coincide for $x\ne2$ (and they are both undefined for $x=-2$), but they are not "mandated" to be equal at $x=2$.

This symptom reflects the difference between

$$\frac{x-2}{x-2}$$ and $$1.$$

0

"Solve $F(x)$" is the wrong terminology. "Simplify $F(x)$" fits better. One solves problems; one solves equations; one evaluates or sometimes simplifies expressions.

$\dfrac 5 0$ is undefined because there is no number $x$ for which $0x = 5.$

But $\dfrac 0 0$ is undefined because there are many numbers $x$ for which $0x=0$, rather than just one such number.

A basic fact of algebra is that if you plug a number into a polynomial in a variable $x$ and get $0$, then $x$ minus that number is a factor of the polynomial. For example, suppose $$ f(x) = x^3 -7x^2 + 5x + 21 $$ so that $$ f(3) = 0. $$ We conclude that $$ x^3 -7x^2 + 5x + 21 = (x-3)(\cdots\cdots\cdots). $$ You still have to do some work to find the other factor, and you get $$ x^3 -5x^2 + 4x + 6 = (x-3)(x^2 - 4x - 7). $$ (Factoring $x^2-2x-2$ further will not concern us for now.)

Now suppose you divide this by another polynomial that is $0$ when $x=3$; for example $g(x) = x^2 - 4x + 3.$ Since $g(3)=0$, you conclude that $g(x) = (x-3)(\cdots\cdots\cdots)$, and when you find the other factor you've got $ g(x) = (x-3)(x-1).$

Now look at $\dfrac{f(x)}{g(x)},$ and see that $\dfrac {f(3)}{g(3)} = \dfrac 0 0$ is undefined.

Now simplify: $$ \frac{f(x)}{g(x)} = \overbrace{ \frac{x^3 - 7x^2 + 5x + 21}{ x^2 - 4x + 3} = \frac{x^2-4x-7}{x-1}}^{\text{when } x\ne3} {} \underbrace{ {} = \frac{-10}{2} = -5}_{\text{when }x=3}. $$

Are we saying that $\dfrac 0 0 = -5$? Are we saying $\dfrac{f(x)}{g(x)} = -5$ when $x=3$?

No, we're not, because one of the "equals" signs is true when $x\ne3$ and the other when $x=3$, so logically we cannot conclude that $\dfrac{f(3)}{g(3)} = -5.$

However, we can conclude that $\dfrac{f(x)}{g(x)}$ can be made as close to $-5$ as desired by making $x$ close enough, but not equal, to $3$. And that is expressed by saying $\dfrac{f(x)}{g(x)}$ approaches $-5$ as $x$ approaches $-3$, or by saying $\dfrac{f(x)}{g(x)} \to -5$ as $x\to-3$, or by saying $\lim\limits_{x\to3} \dfrac{f(x)}{g(x)} = -5.$

Michael Hardy
  • 1