Answer
See graph and explanations.
Work Step by Step
Step 1. Given $y=f(x)=2x^4-4x^2+1$, we have $y'=8x^3-8x=8x(x^2-1)=8x(x+1)(x-1)$ and $y''=24x^2-8=24(x^2-\frac{1}{3})=24(x+\frac{\sqrt 3}{3})((x-\frac{\sqrt 3}{3})$
Step 2. The inflection points can be found at zeros of $y''$ where there is a sign (concavity) change. We have $..(+)..(-\frac{\sqrt 3}{3})..(-)..(\frac{\sqrt 3}{3})..(+)..$ and we can identify two inflection points at $x=-\frac{\sqrt 3}{3}$ and $x=\frac{\sqrt 3}{3}$.
Step 3. The extrema can be found at zeros of $y'$ with sign changes in $y'$: $..(-)..(-1)..(+)..(0)..(-)..(1)..(+)..$ and we can identify a local maximum at $(0,1)$ and two local minima at $(-1,-1),(1,-1)$.
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.
