-1
$\begingroup$

taking an example

The function $f(x)=\dfrac{x^2-16}{x-4}$ is not defined at $x=4$ but on simplifying it to $x+4$ it gets defined at $x=4$. So does the factorizing and simplification changes the function?

Saying it in different way
Multiplying $x+a$ with $\dfrac{x-n}{x-n}$ (which is basically $1$) makes it undefined at some value $n$.
This means multiplying by 1 changes the function. Is this true? Because the graph of the function will get a hole in it but on other hand we always write $x\cdot 1=x$.

  • 1
    Nilabro Saha gave a good answer already. The concept you are looking for might be "extending the function continuously". That way you can assign a value at $x=4$ and the function is still continuous. – 2017-02-26

1 Answers 1

2

Sure the function

$$ f(x) = \frac{x^2 - 16}{x - 4} $$

is not defined at $x = 4$. When you "simplify" it to $f(x) = x + 4$, you do it under the consideration that $x \ne 4$.

And $\frac{x - n}{x - n} = 1$ only for $x \ne n$. So when you multiply $x + a$ by $1 = \frac{x - n}{x - n}$ you do it considering $x \ne n$.

Suffices? Guess they don't teach this in high school.

  • 0
    Then saying f(x)=8 at x=4 is not correct because we have simplified by assuming that x is not equal to 4 – 2017-02-26
  • 0
    Obviously. You're slowly getting it. – 2017-02-26
  • 0
    So functions f(x)=x^2-16/x-4 and g(x)= x+4 is exactly same or a little different – 2017-02-26
  • 0
    $f(x) = g(x)$ for $x \ne 4$. – 2017-02-26