Determine the intervals of monotonicity of the function y

Determine the intervals of monotonicity of the function $y = x+\bigl|\sin2x\bigr|$.

Step 1: Remove the absolute value

We have $\sin2x\ge0$ exactly when $2x\in\bigl[2k\pi,\bigl(2k+1\bigr)\pi\bigr]$, that is when $x\in\left[k\pi,\,k\pi+\dfrac{\pi}{2}\right]$; otherwise $\sin2x\lt 0$. Hence

\[ y = \begin{cases} x+\sin2x, & x\in\left[k\pi,\;k\pi+\dfrac{\pi}{2}\right],\\[6pt] x-\sin2x, & x\in\left(k\pi+\dfrac{\pi}{2},\;\bigl(k+1\bigr)\pi\right). \end{cases} \]

The whole picture repeats with period $\pi$, since $y\bigl(x+\pi\bigr) = y(x)+\pi$.

Step 2: The derivative on each piece

\[ y' = 1+2\cos2x \quad\text{on}\quad \left(k\pi,\;k\pi+\frac{\pi}{2}\right), \tag{1} \]

\[ y' = 1-2\cos2x \quad\text{on}\quad \left(k\pi+\frac{\pi}{2},\;\bigl(k+1\bigr)\pi\right). \tag{2} \]

At the junctions $x = k\pi$ and $x = k\pi+\dfrac{\pi}{2}$, where $\sin2x = 0$, the function is continuous but has corners: the one-sided derivatives are $1\mp2$, so they differ by $4$.

Step 3: First piece: where is $1+2\cos2x$ positive?

Setting $y' = 0$ gives $\cos2x = -\dfrac12$. On $\left(k\pi,\,k\pi+\dfrac{\pi}{2}\right)$ the angle $2x$ runs through $\bigl(2k\pi,\,2k\pi+\pi\bigr)$, where the only solution is $2x = 2k\pi+\dfrac{2\pi}{3}$, i.e. $x = k\pi+\dfrac{\pi}{3}$. Therefore

\[ \left(k\pi,\;k\pi+\frac{\pi}{3}\right):\ \cos2x\gt -\tfrac12 \Rightarrow y'\gt 0 \ (\nearrow); \qquad \left(k\pi+\frac{\pi}{3},\;k\pi+\frac{\pi}{2}\right):\ y'\lt 0 \ (\searrow). \]

Step 4: Second piece: where is $1-2\cos2x$ positive?

Now $y' = 0$ means $\cos2x = \dfrac12$. On $\left(k\pi+\dfrac{\pi}{2},\,\bigl(k+1\bigr)\pi\right)$ the angle $2x$ runs through $\bigl(2k\pi+\pi,\,2k\pi+2\pi\bigr)$, where the only solution is $2x = 2k\pi+\dfrac{5\pi}{3}$, i.e. $x = k\pi+\dfrac{5\pi}{6}$. Therefore

\[ \left(k\pi+\frac{\pi}{2},\;k\pi+\frac{5\pi}{6}\right):\ \cos2x\lt \tfrac12 \Rightarrow y'\gt 0 \ (\nearrow); \qquad \left(k\pi+\frac{5\pi}{6},\;\bigl(k+1\bigr)\pi\right):\ y'\lt 0 \ (\searrow). \]

Step 5: Verification

\[y'(0.8) = 0.9416\gt 0; \quad y'(1.4) = -0.8844\lt 0; \quad y'(1.7) = 2.9336\gt 0; \quad y'(3.0) = -0.9203\lt 0 \;\;\checkmark\]

\[y\!\left(\tfrac{\pi}{3}\right) = 1.9132 \gt y\!\left(\tfrac{\pi}{2}\right) = 1.5708; \qquad y\!\left(\tfrac{5\pi}{6}\right) = 3.4840 \gt y\bigl(\pi\bigr) = 3.1416 \;\;\checkmark\]

\[\text{Corners: } y'\!\left(\tfrac{\pi}{2}^{-}\right) = 1+2\cos\pi = -1, \quad y'\!\left(\tfrac{\pi}{2}^{+}\right) = 1-2\cos\pi = 3 \;\;\checkmark\]