Question

whats a good way to explain my 12 year old niece :
why \(f(x) = \dfrac{x^2-x}{x-1} = x\) is not defined when \(x=1\) ?

Answer 1

well if x=1\[\frac{ 1^{2}-1 }{ 1-1 }\]
\[\frac{ 1-1 }{ 1-1 }\]

Answer 2

in short terms you're subtracting 1 by itself

Answer 3

I was actually thinking of how to explain this the other day. This is what I came up with. So x/y is basically saying you have x and you put it into y groups. Well, if you start with x=0, then you have 0 to start with. If you have 0 of something, it can't be separated into groups.

Answer 4

explain x^2= x*x if x=1 you're multiplying 1 by itself then subtracting by itself cancelling itself out of the equation

Answer 5

because if you divide by zero your head will turn around like linda blaire in the exorcist

Answer 6

i wanna try that wanna see my head turn around

Answer 7

also because \(f(x) = \dfrac{x^2-x}{x-1} = x\) is false \[f(x) = \dfrac{x^2-x}{x-1} = x \text{ only if }x\neq 1\] is true

Answer 8

Oh yes very good points ! I think it boils down to 0/0 being undefined... but im not sure if il be able to convince her why 0/0 is undefined as I still need to do study more on how to teach her in an efficient+simple way

Answer 9

the domain of \(f(x)=\frac{x^2-x}{x-1}\) doesn't include 1 so when you factor and cancel you are assuming \(x\) is not \(1\) to begin with

Answer 10

my teacher back in 7th grade explained undefined by having us make a line in our seats with our body if it's a horizontal line slope is undefined

Answer 11

`explain x^2= x*x if x=1 you're multiplying 1 by itself then subtracting by itself cancelling itself out of the equation`

we're loosing the variable here ? im thinking of a similar similar thing - change it to multiplication equation and interpret :

\[\large f(x) = \dfrac{x^2-x}{x-1} \implies f(x)(x-1)=x^2-x\]

Answer 12

you're not losing a variable
x=1
x^2=x*x

Answer 13

you're expanding it

Answer 14

how i would explain this to her is tell her to solve that and give you the slope if she cant draw a horizontal line then tell her to find the slope

Answer 15

undefined=0 slope

Answer 16

that makes sense :) before giving her geometric explanation, is it a good idea start with :

\(ab = cb \implies a=c \) only when \(b\ne 0\)

and give her few examples may be :
because \(2*0 = 3*0 \)
etc..

Answer 17

if above looks good, i can follow up with the actual concept :
\(\large f(x)\color{red}{(x-1)}=x\color{Red}{(x-1)} \implies f(x) = x\) only when
\(\color{red}{(x-1)\ne 0}\)

looks bit mechanical to me, but should be okay i guess... this undefined concept itself is very confusing >.<

Answer 18

well dont confuse the kid >.< explain what slope is and how to find it in point slope form and in a graph how slope= rise over run

Answer 19

could you elaborate please - i kindof got it, but still somewhat unclear. how to use slope to explain this ?

Answer 20

ehh thought you were talking about point slope form my bad >.< forgot all variable were x's

Answer 21

I would say it starts with the definition of division.
We can say \(
\frac{a}{\color{red}b}
\) is undefined when \(\color{red}b=0\).
This means that when \(\color{red}{x-1}=0\) then \(
\frac{x^2-x}{\color{red}{x-1}}
\) is undefined.

Therefore, we must concluded that: \[
x\quad \frac{x^2-x}{x-1} = x
\]is not a true statement. At least it is not true when \(x=1\).

To be completely accurate, we would have to say: \[
\frac{x^2-x}{x-1} = \begin{cases}
x &x\neq 1\\
\text{undefined} &x=1
\end{cases}
\]



This helps understand the purpose of limits.
We could also say: \[
\lim_{x\to a}\frac{x^2-x}{x-1} = a
\]

In fact, due to properties of limits, in general we could say: \[
\left[\lim_{x\to a} \begin{cases}
f(x)&x\neq c\\
\ldots &x=c
\end{cases} \right]= f(a)
\]because limits will smooth out single point discontinuities.

Answer 22

nice, to the point :) thanks a lot everyone !

Answer 23

ur niece is amazing xD
im trying to teach my 13 yrs old nephew
but we are still stuck on area of square/circl -.-