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Please check my method and also if I have solved the following problem correctly:

Problem: $f(x) = |x - \frac12| + |x + \frac12|$

If $x = -1$, then:

$f(-1) = |-1 - \frac12| + |-1 + \frac12|$

From the definition of $|x|$ we see that, $$|a-b| = \begin{cases} a - b & \text{if }a\geq b \\ b - a & \text{if }b \geq a\end{cases}$$ It is the order property for reals numbers that if $a$ is to the left of $b$ on the number line, then $a<b$ and vice versa.

Here, $-1$ is to the left of $-\frac12$, and also $-1$ is to the left of $\frac12$. Therefore, $-1<-\frac12$, and also $-1<\frac12$. So,

$f(-1) = \frac12 -(-1) + \frac12 -(-1)$

$f(-1) = \frac12 +1 + \frac12 +1$

$f(-1) = \frac12 +\frac12 + \frac21 $

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

$f(-1) = \frac{4} {2}$

$f(-1) = 2$

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    $\begingroup$ Why bother with the rule about $|a-b|$? Why not just perform the sums inside the absolute values and work with the more basic definition? That is, $f(-1)=|-\tfrac{3}{2}|+|-\tfrac{1}{2}|=\tfrac{3}{2}+\tfrac{1}{2}=2$? $\endgroup$
    – Michael Grant
    Apr 25, 2013 at 13:11
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    $\begingroup$ It does look like you solved the problem correctly, mind you, I just think this approach is unnecessarily complex. $\endgroup$
    – Michael Grant
    Apr 25, 2013 at 13:11
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    $\begingroup$ Your calculations are correct, but for the future, just calculate whatever inside the absolute value, and at the end remove minus sign, if there's one. $\endgroup$
    – Kaster
    Apr 25, 2013 at 13:12
  • $\begingroup$ I just wanted to tell that the method behind my calculations is the one I've mentioned. Otherwise, I'm not going to use it again. Someone suggested me, "You should also express the method if you want us to check your calculations." $\endgroup$
    – Samama Fahim
    Apr 25, 2013 at 13:15

2 Answers 2

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Actually it looks like you made two mistakes, which happened to cancel each other out to lead to the right answer. The first mistake: $$\left|-1+\frac{1}{2}\right|\neq \frac{1}{2}-(-1)$$ Instead, it should be $|-1+\tfrac{1}{2}|=-\frac{1}{2}-(-1)$. The second mistake: $$\frac{1}{2}+\frac{1}{2}+\frac{2}{1}\neq \frac{1+1+2}{2}.$$ Instead, this would be $\tfrac{1+1+4}{2}$. This is in turn equal to $3$, which is the wrong answer, but it follows from the previous mistake.

I would say that the first mistake supports our unanimous contention that you made your approach far more complex than necessary. You should have simply evaluated the sums/differences inside each absolute value, and then dropped the signs. This rule about $|a-b|$ is unnecessary.

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If the problem is to evaluate $f(-1)$, you get the correct answer but it looks like you got there incorrectly. In your first equation $f(-1) = -\frac12 -(-1) + \frac12 -(-1)$ it looks like you are claiming $|-\frac 12 -1|=\frac 12$ (the first half) and $|\frac 12 -1|=\frac 32$. Both of these are wrong. When you change the sign in the absolute value bars, you need to change the sign of the $\frac 12$ as well. As Michael Grant says, you should evaluate what is inside the absolute value bars and then fix the sign.

If you had been asked for $f(0)$ it looks like you would get $f(0)=-\frac 12+0 +\frac 12 + 0=0$ when you should get $f(0)=+\frac 12 + \frac 12=1$

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