Absolute Value Problem, Solution and Method

Question

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 $

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

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

$f(-1) = 2$

Answer 1

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.

Answer 2

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 $f(0)=+\frac 12 + \frac 12=1$