Regarding functions how can simplifying a function change it's domain

[Marmoteer]

for example
if f(x) = \frac{x^{2} - 1}{x+1}

they by factoring:

f(x) = \frac{x^{2} - 1}{x+1}<br /> = \frac{(x + 1)(x - 1)}{x+1} <br /> = x - 1

thus the simplified version is x - 1

let's say g(x) = simplified f(x)

g(x) = x - 1

meaning g(x) = f(x)

but

g(-1) ≠ f(-1)

am I missing something?

 

[BloodyFrozen]

$f(x) = \frac {x^2-1}{x+1}$
$f(0) = \frac {0^2-1}{0+1}$
$f(0) = -1$

$g(x) = x-1$
$g(0) = 0-1$
$g(0) =-1$
The two functions act the same except at x = -1 at which $f(x)$ has a hole at -1.

 

[Marmoteer]

Oops Sorry I meant to write g(-1) ≠ f(-1)

 

[Stephen Tashi]

how can simplifying a function change it's domain

Symbolic manipulations in math are an abbreviation for thought, not a substitute for it. If you "simplify" an expression without thinking about it, you can make a faulty deduction. The correct deduction is

\frac{(x+1) (x-1)}{x+1} = x-1 except when x = -1.

There may be types of math where you only manipulate symbols and never use ordinary language, but algebra is not one of them.

 

[HallsofIvy]

for example
if f(x) = \frac{x^{2} - 1}{x+1}

they by factoring:

f(x) = \frac{x^{2} - 1}{x+1}<br /> = \frac{(x + 1)(x - 1)}{x+1} <br /> = x - 1

thus the simplified version is x - 1

let's say g(x) = simplified f(x)

g(x) = x - 1

meaning g(x) = f(x)

What you are "missing" is that this is not true. What is true is that f(x)= g(x) for all x except x= -1. f and g are NOT the same function.

but

g(-1) ≠ f(-1)

am I missing something?

 

[mesa]

Symbolic manipulations in math are an abbreviation for thought, not a substitute for it. If you "simplify" an expression without thinking about it, you can make a faulty deduction. The correct deduction is

\frac{(x+1) (x-1)}{x+1} = x-1 except when x = -1.

There may be types of math where you only manipulate symbols and never use ordinary language, but algebra is not one of them.

Is there an easy way to spot a problem like this, or is it something you just get better at over time?

 

[HallsofIvy]

You should have learned long ago that you cannot divide by 0 so that the division in
\frac{(x- 1)(x+1)}{x+1}= x- 1
is only valid when x+ 1\ne 0 which is only true when x\ne -1.

 

[Stephen Tashi]

Is there an easy way to spot a problem like this, or is it something you just get better at over time?

There are common traps, like expessions that "don't work" for some values of variables because they involve division by zero or square roots of negative numbers etc.

You'll get better at spotting the common traps over time if you become suspicious of any form of calculation. Don't trust. Don't believe.

 

[mesa]

There are common traps, like expessions that "don't work" for some values of variables because they involve division by zero or square roots of negative numbers etc.

You'll get better at spotting the common traps over time if you become suspicious of any form of calculation. Don't trust. Don't believe.

The dividing by zero makes a lot of sense, I was surprised to see it still apllied even when the numerator would also equal 0 hence 0/0. I will certainly be mindful of this and irrational numbers in the future, thanks for the tip!