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

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

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Answer 1 · 2017-12-27T11:12:06

$f (x) = - 2 x$ $f(x) = -2x$ for $x \geq - \frac{3}{2}$ $x>=-3/2$ and $2 x + 6$ $2x+6$ for $x < - \frac{3}{2}$ $x<-3/2$ .

Explanation:

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

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

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

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

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

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

For $x \geq - \frac{3}{2}$ $x>=-3/2$ : $f (x) = 3 - (2 x + 3) = - 2 x$ $f(x) = 3- (2x+3) = -2x$
For $x < - \frac{3}{2}$ $x<-3/2$ : $f (x) = 3 - (- 2 x - 3) = 2 x + 6$ $f(x) = 3 - (-2x-3) =2x+6$

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

$f (x) = - 2 x$ $f(x) = -2x$ for $x \geq - \frac{3}{2}$ $x>=-3/2$ and $2 x + 6$ $2x+6$ for $x < - \frac{3}{2}$ $x<-3/2$ .

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

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

1 Answer

Explanation:

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