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While practicing problems on functions, I am getting a lot of incorrect answers from the given answers. I am a little confused by this as the problems seem simple else I haven't really understood the underlying principles. Can you guys help me figure if my answers are correct?

  1. $f:x \rightarrow |x| + |x-2|$ ,My Answer: $[2,\infty)$ ,Given:$[0,\infty)$

  2. $f:x \rightarrow |x+1| + |x-3|$ ,My Answer: $[4,\infty)$ ,Given: $[0,\infty)$

  3. $f:x \rightarrow x + |x-4|$ ,My Answer: $\mathbb{R}$ ,Given: $[4,\infty)$

Thanks for your help.

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    Thanks, Maple sounds interesting. Will take a look.2011-06-20

4 Answers 4

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Usually it's easier to write explicitly the expressions of the function, when it involves absolute values. Use the property $|a|=\begin{cases}a & a \geq 0 \\ -a & a<0\end{cases}$

  1. $f(x)=\begin{cases}2-2x & x < 0 \\ 2 & x \in [0,2] \\ 2x-2 & x >2 \end{cases}$. From here it's easy to see the image of $f$.
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Already posted are several practical algebraic methods for dealing with the problems. What follows is less practical, but may be a useful idea in other settings.

In general, $|x-a|$ is the distance between $x$ and $a$. Now let's look for example at Problem $2$.

We are looking at $|x-(-1)| + |x-3|$. Draw a "number line" and put dots at $-1$ and at $3$. Then $|x-(-1)| + |x-3|$ is the sum of the distances from $x$ to the two numbers $-1$ and $3$.

This sum is clearly $4$ if $x$ is between $-1$ and $3$. Now let us start at $x=3$ and imagine $x$ moving slowly to the right. Then the sum of the distances is at first $4$, and then steadily increases, becoming large after a while. Now start at $-1$ and move slowly to the left. Again, the sum of the distances is at first $4$, and steadily increases.

It follows that the correct answer for ($2$) is $[4,\infty)$.

The idea also works for Problem $3$, but becomes geometrically less natural. Start at $x=4$ and move steadily to the right. Then at first $x+|x-4|$ is $4$, then steadily increases. Now start at $4$ and move steadily to the left. Every increase in the quantity $|x-4|$ is exactly compensated by a decrease in $x$, so the sum $x+|x-4|$ remains unchanged at $4$.

It follows that the correct answer for ($3$) is $[4,\infty)$.

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    Wow! This is a very cool way of thinking. Thank you!2011-06-20
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Hint: Use the Triangle Inequality, $ |x| + |y| \ge |x - y| $ (for 1 and 2)

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    Thank you for that, corrected.2011-06-20
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For 3), if $x \leq 4$ then $ x + |x-4| = x + (4-x) = 4, $ and if $x > 4$ then $ x + |x-4| > x. $ So the given answer is correct for 3), that is the range is $[4,\infty)$, not $\mathbb{R}$.

For 1) and 2), your answers are correct. For 1), it is useful to consider the cases $x \leq 0$, $0 \leq x \leq 2$, and $x \geq 2$, and for 2) the cases $x \leq -1$, $-1 \leq x \leq 3$, and $x \geq 3$.

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    I had a feeling I might have mixed up. Thanks for your help.2011-06-20