0
$\begingroup$

When breaking an absolute value function such as $y = |2-x|$ into parts for a piece-wise function, would $2-x$ be the piece for $x < 2$, and $x-2$ be the piece for $x ≥ 2?$

If so, is it because $y = |2-x|$ is the same as $|-x+2| = |-(x-2)| = x-2?$ In other words, is $y = |2-x|$ already given in its negative form, so there would be no need to repeat the process of doing $-(2-x)$ when finding the piece for $x < 2?$

For example, for $y = |x|$, in order to find the piece for $x < 0$, you would have to apply the negative sign outside of $|x|$ and then becoming $-|x|$ then $-(x)$ and then $-x$ . But when finding the piece for $x < 2$ of $y = |2-x|$, is applying the negative sign outside of the absolute value bars not needed since $|2-x|$ is already the piece for $x < 2$, since $$|2-x| = |-x+2| = |-(x-2)| = -|x-2|= -(x-2) = 2-x?$$

It's kind of difficult to explain but is this the correct reason why $|2-x|$ is still $2-x$ when $x < 2$ instead of $-(2-x)$ in the piece-wise function?


EDIT1: When I say "$y = |2-x|$ already given in its negative form" I mean that there's no need to apply the negative sign outside of the absolute value brackets to obtain the piece for $x < 2$. For example, with something like $y = |x|$, you would have to apply the negative signs outside to obtain $-x$, which is the piece for $x<0$.

$\endgroup$
  • $\begingroup$ Placing a negative sign outside the absolute value bars changes the sign of the expression unless the quantity inside the absolute value bars is equal to zero. When you write an absolute value in piecewise form, what matters is the sign of the quantity inside the absolute value bars. $\endgroup$ – N. F. Taussig Sep 29 '19 at 7:50
  • $\begingroup$ So basically |2-x| is already in its negative form, since there already was a negative sign applied to the quantity inside the absolute value bars? E.g. |2-x| = |-(x-2)| $\endgroup$ – Patrick Pichart Sep 29 '19 at 7:53
  • $\begingroup$ I know y = |x| and y = |-x| are equivalent when graphed, but what's the difference between them? $\endgroup$ – Patrick Pichart Sep 29 '19 at 7:57
  • $\begingroup$ Note that $|2 - x| = |(-1)(x - 2)| = |-1||x -2| = 1|x - 2| = |x - 2|$ and that $|-x| = |(-1)x| = |-1||x| = |x|$. $\endgroup$ – N. F. Taussig Sep 29 '19 at 8:01
  • $\begingroup$ My question is why does y = |-x| remain the same as in what's in the inside, when finding the piece for x<0? For example, for x < 0 for y = |x|, you have to apply the negative signs on the inside to get -x, but for y = |-x|, you just leave it as is. $\endgroup$ – Patrick Pichart Sep 29 '19 at 8:05
0
$\begingroup$

The absolute value of a real number is its distance from $0$ on the real number line.

For instance, $|5| = 5$ since $5$ is five units from $0$, and $|-4| = 4$ since $-4$ is four units from $0$.

If $x \geq 0$, the distance of $x$ from $0$ is just $x$. If $x < 0$, the distance of $x$ from $0$ is $-x$. This gives us the piecewise definition $$|x| = \begin{cases} x & \text{if $x \geq 0$}\\ -x & \text{if $x < 0$} \end{cases} $$ that you stated.

Let's look at $|2 - x|$. Observe that \begin{align*} 2 - x & \geq 0\\ 2 & \geq x \end{align*} Hence, the quantity inside the absolute value bars is nonnegative when $x \leq 2$. It is negative otherwise. Thus, \begin{align*} |2 - x| & = \begin{cases} 2 - x & \text{if $x \leq 2$}\\ -(2 - x) & \text{if $x > 2$} \end{cases}\\ & = \begin{cases} 2 - x & \text{if $x \leq 2$}\\ -2 + x & \text{if $x < 2$} \end{cases} \end{align*} Notice that the piecewise definition of $|2 - x|$ is determined by the sign of the quantity $2 - x$. When $2 - x \geq 0$, $|2 - x| = 2 - x$. On the other hand, if $2 - x < 0$, $|2 - x| = -(2 - x) = -2 + x = x - 2 > 0$.

Addendum

I wish to call your attention to errors you made.

Since the absolute value of a number cannot be negative, $$|x - 2| = x - 2$$ is only true if $x - 2 \geq 0$, which occurs when $x \geq 2$. If $x < 2$, then $x - 2 < 0$, so $|x - 2| = -(x - 2) = -x + 2 = 2 - x > 0$.

The statement $$|-(x - 2)| = -|x - 2|$$ is only true if $x - 2 = 0$. If $x \neq 2$, the expression on the left-hand side is positive, while the quantity on the right-hand side is negative, making the statement false. What you should have written is $$|-(x - 2)| = |(-1)(x - 2)| = |-1||x - 2| = 1|x - 2| = |x - 2|$$

$\endgroup$
  • 1
    $\begingroup$ Ahh I finally understand now! Thanks so much! $\endgroup$ – Patrick Pichart Sep 29 '19 at 8:37
  • $\begingroup$ My understanding now is that an absolute value of a number is the distance from 0 on the number line, so the sign will not matter since it is only based on the magnitude; in other words, it's always positive. For the piece-wise definition of an absolute value function, it is itself if it is either positive or 0 and is the opposite of itself when negative, so a negative needs to be multiplied to that negative number to make it positive. $\endgroup$ – Patrick Pichart Sep 29 '19 at 11:24
  • $\begingroup$ And writing the inequality of x≥a or x<a and then isolating for x will give the new inequality of what conditions the the absolute value function will be positive and what conditions it will be negative. $\endgroup$ – Patrick Pichart Sep 29 '19 at 11:28
  • $\begingroup$ The absolute value of a number is always nonnegative. When the quantity inside the absolute value is nonnegative, it is equal to its absolute value. When the quantity inside the absolute value is negative, we must multiply it by $-1$ to obtain its absolute value. $\endgroup$ – N. F. Taussig Sep 29 '19 at 11:32
  • $\begingroup$ In the most simplest way I'd like to explain this is, an absolute value simply outputs positive values. So for |2−x|, it results in the same expression as what's in the inside of the bars when x ≤ 2, since when subtracting any number ≤ 2, it will always be postive or zero. But for any number > 2, 2 subtracted by that number will always result in a negative a number inside the abs value bar. So since, an absolute value makes everything positive, it has to multiply anything that's negative inside by -1 to make it positive; thus -(2-x) would be the expression for x>2. Right? $\endgroup$ – Patrick Pichart Sep 29 '19 at 11:52

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.