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?
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.