1
$\begingroup$

Is there an easier way to (I am aware of my poor translation, but am not familiar with english terminology; however, I think you will understand.) "Reduce the following fraction:"

$\dfrac{x | x-1 |-x+1}{x^2-2|x|+1}$

besides doing numerous situations with absolute values (in this case, 4 situations); in this case I'd make:
i.) $x - 1 \ge 0$, $\space x \le 0$

ii.) $x - 1 \le 0$, $\space x \ge 0$

iii.) $x - 1 \ge 0$, $\space x \ge 0$

iv.) $x - 1 \le 0$, $\space x \le 0$

Then solve each pair's cross-section of the two solutions...

Second thing I'm not certain about are the "$\ge$" and "$\le$" signs. Do I have to put on both equations in the pair the equal sign "$=$", or just "$>$" / "$<$". If it's not clear, do I have to make every like:

i.) $x - 1 \ge 0$, $\space x \le 0$

ii.) $x - 1 \le 0$, $\space x \ge 0$

or

i.) $x - 1 \ge 0$, $\space x < 0$

ii.) $x - 1 \le 0$, $\space x > 0$

Please explain when to put the "$\le$" and "$\ge$".

EDIT: Basically, what I wanted to know is, if there is any kind of faster - automated - way of solving the question. The answers you provided matched my opinion and praxis. The thing here is, I know a faster way of solving these, it includes a, so called "table" (with 'null-points', something like RecklessReckoner advised), if someone is interested I'll make a photo of it. It makes solving these totally automated that you don't have to think about the actual problem and/or understand it, that's why I tried to solve these traditionally to understand it - behind the scenes. Once again, thanks for your answers.

$\endgroup$
  • 1
    $\begingroup$ I do not see anything much shorter. There will be $3$ cases: $x\gt 1$; $0\le x\lt 1$; $x\lt 0$. $\endgroup$ – André Nicolas Aug 31 '13 at 3:09
  • $\begingroup$ You can toss out one of those four cases (in the OP) right away. $\endgroup$ – aschepler Aug 31 '13 at 3:09
1
$\begingroup$

It takes thinking. When you split into cases, generally you have something like $|x-a|$, which will be $x-a$ if $x \ge a$ and $a-x$ if $a \ge x$. If $x=a$ it doesn't matter which direction you use, so generally you use the sign with the one with the equals because you don't want to exclude a potential solution. The point is "do you know it isn't equal-otherwise include the equals. Then you need to check the solutions you find in the original equation because you may have introduced extraneous solutions.

$\endgroup$
1
$\begingroup$

I generally tell students to approach functions containing expressions in absolute value brackets this way. The expression $ \ \vert u(x) \vert \ $ in brackets introduces "special points" at the values of $ \ x \ $ where $ \ u(x) = 0 \ $. To either side of these points will be intervals where the expression is read as $ \ u(x) \ $ if $ \ u(x) > 0 \ $ and $ \ -u(x) \ $ if $ \ u(x) < 0 \ $ . This reduces the matter of interpretation to one of reading the behavior of the function in each interval, rather than a "logic puzzle" requiring consideration of cases based on a collection of "binary switches", so to speak.

As you and the commenters have noted, there are two "special values" at $ \ x = 0 \ $ and $ \ x = 1 \ $ , so, all told, there are five possible "readings" of the rational function in question:

for $ \ x < 0 \ , $ where $ \ \vert x \vert = -x \ $ and $ \ \vert x-1 \vert = 1-x \ ; $

for $ \ x = 0 \ , $ where $ \ \vert x \vert = 0 \ $ and $ \ \vert x-1 \vert = 1 \ ; $

for $ \ 0 < x < 1 \ , $ where $ \ \vert x \vert = x \ $ and $ \ \vert x-1 \vert = 1-x \ $ ;

for $ \ x = 1 \ , $ where $ \ \vert x \vert = 1 \ $ and $ \ \vert x-1 \vert = 0 \ $ ; and

for $ \ x > 1 \ , $ where $ \ \vert x \vert = x \ $ and $ \ \vert x-1 \vert = x - 1 \ $ .

(This last produces an especially simple result...)

ADDENDUM: That being said, this particular rational function has an additional complication at $ \ x = -1 \ , $ owing to the form of the denominator function. This, it seems, is not evident from looking at the absolute value expressions alone...

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.