# How do you write f(x) = 3 - |2x + 3| into piecewise functions?

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#### Explanation

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#### Explanation:

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1
Dec 27, 2017

$f \left(x\right) = - 2 x$ for $x \ge - \frac{3}{2}$ and $2 x + 6$ for $x < - \frac{3}{2}$.

#### Explanation:

It's really the absolute value you need to worry about.

If you're rewriting absolute values of the form $| u |$ start with finding the zero of $u$: $2 x + 3 = 0 \setminus \rightarrow x = - \frac{3}{2}$.

Now figure out where $2 x + 3$ is positive and negative.

By inspection (or picking values to substitute) $2 x + 3 < 0$ for $x < - \frac{3}{2}$ and $2 x + 3 \ge 0$ for $x \ge - \frac{3}{2}$.

In general $| u | = u$ when $u \ge 0$ and $| u | = - u$ when $u < 0$, so now we have:

$| 2 x + 3 | = 2 x + 3$ for $x \ge - \frac{3}{2}$ and $| 2 x + 3 | = - 2 x - 3$ for $x < - \frac{3}{2}$. We make those substitutions.

For $x \ge - \frac{3}{2}$: $f \left(x\right) = 3 - \left(2 x + 3\right) = - 2 x$
For $x < - \frac{3}{2}$: $f \left(x\right) = 3 - \left(- 2 x - 3\right) = 2 x + 6$

So that's our function. (Not sure how I'd write a piecewise function on here using the normal notation.)

$f \left(x\right) = - 2 x$ for $x \ge - \frac{3}{2}$ and $2 x + 6$ for $x < - \frac{3}{2}$.

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