I'm not sure how to express this function in piecewise form without using absolute values:
$$
f(x) = 3|x-2| – |x+1|$$
I know how to do it when there is just one absolute value, such as:
$$g(x) = 3+|2x-5|$$
$$
g(x)=
\begin{cases} 2x-2& \text{; }x\ge\frac52\\
8-2x&\text{; }x<\frac52 \end{cases}
$$
To express $g(x)$ in piecewise form, I made 2 cases. Case 1 positive and Case 2 negative.
But I can't exactly do that for this problem [$f(x)$]… Could someone tell me how to proceed?
EDIT: Thank you for all the help!
Best Answer
The only places where we need to break the function into "pieces" are those points where the expression inside each absolute value becomes zero. This occurs at $x = -1$ and $x = 2$.
Accordingly, we consider what the function looks like on each of the intervals $(-\infty,-1)$, $(-1,2)$, and $(2,\infty)$.
For $x < -1$, we have $x + 1 < 0$ and so $|x+1| = -(x+1)$. Similarly, $x - 2 < x + 1 < 0$ and so $|x-2| = -(x-2)$. So on the interval $(-\infty,-1)$ the function can be simplified as $f(x) = -3(x-2) +(x+1) = -2x+7$.
Thus far, we have established
$$f(x) = \begin{cases} -2x + 7 & \text{if }x < -1 \\ ??? & \text{if }-1 \leq x \leq 2 \\ ??? & \text{if }x > 2 \\ \end{cases}$$
You can handle the other intervals in exactly the same way.