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Consider the function $$f(x) = 1\dfrac{1}{2} - 3\sin \left(\dfrac{1}{2}x \right). $$

I need to find the absolute of this function, which to my eye would just be

$$ f(x) = 1\dfrac{1}{2} + 3\sin \left(\dfrac{1}{2}x \right), $$ but that's incorrect.

Why is this incorrect and how can you find the absolute value of such functions?

Kaish
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Absolute
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  • Changing the sign of the term $3 \sin(\frac{1}{2} x)$ just turns the whole function "upside down": the positive parts become negative and vice versa. You want to leave the positive parts alone and only change the sign of the negative portions of $\sin(\frac{1}{2} x)$. This would be $\frac{3}{2} + 3 \cdot |\sin(\frac{1}{2} x)|$. I'm not clear yet as to what you want to do with this... – colormegone Apr 18 '13 at 21:55
  • @RecklessReckoner: Changing the sign that way certainly does not turn the whole function upside down. For example, $f(0)$ is positive, and if you change the sign that way, you'll still get a positive number when $x=0$. – Cameron Buie Apr 18 '13 at 22:01
  • I'm referring to the single trigonometric term, which is where the "sign flip" was made. Yes, for the function, one would need to locate the intervals where $f(x) < 0$ and turn those portions of the function over. – colormegone Apr 18 '13 at 22:03
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    The absolute value of this function is $g(x)=|f(x)|=| 1 \frac{1}{2}−3 \sin(\frac{1}{2}x)|$ :) – thesamet Apr 18 '13 at 22:12

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Rewrite as $$f(x)=\frac32\left(1-2\sin\frac12x\right).$$ Since $\frac32\ge0$, then $|f(x)|=f(x)$ whenever $1-2\sin\frac12x\ge 0$ and $|f(x)|=-f(x)$ otherwise.

Cameron Buie
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