Answer
See graph and explanations.
Work Step by Step
Step 1. Given $y=f(x)=x\ cos(x), 0\leq x\leq 2\pi$, we have $y'=cos(x)-x\ sin(x)$ and $y''=-sin(x)-sin(x)-x\ cos(x)=-2sin(x)-x\ cos(x)$
Step 2. The inflection points can be found at zeros of $y''$, we have $-2sin(x)-x\ cos(x)=0$ which gives $x=0, 2.289,5.087$ in the domain. Check the sign (concavity) changes of $y''$, we have $(0)..(-)..(2.289)..(+)..(5.087)..(-)..(2\pi)$ and we can identify two inflection points at $x=2.2893$ and $x=5.087$.
Step 3. The extrema can be found at zeros of $y'$, we have $cos(x)-x\ sin(x)=0$ which gives $x=0.86, 3.426$ in the domain. Check the sign changes in $y'$; we have $(0)..(+)..(0.86)..(-)..(3.426)..(+)..(2\pi)$ and we can identify two local maxima at $(0.86,0.561), (2\pi,2\pi)$ and two local minima at $(0,0),(3.426,-3.288)$.
Step 4. Graph the function and its derivatives as shown in the figure.
Step 5. The intersection of $y'$ with the x-axis indicates possible extrema positions.
Step 6. The intersection of $y''$ with the x-axis indicates possible inflection points.
Step 7. Positive $y'$ indicates increasing region and negative $y'$ indicates decreasing region of the function.
Step 8. Positive $y''$ indicates concave up region and negative $y''$ indicates concave down region of the function.
