$$f(x)=x^2(x-1)^2$$
Sketch the graph of $f(x)$ and label all the critical points $C(x, y)$ and inflection points $I(x, y)$ on the graph.
So I've got the critical points at $$C(0,0), C(1/2,1/16), C(1,0)$$
And then I've calculated the second derivate of $f(x)$ Which is $$f''(x) = 1/2 ± (√5/3)/2$$ $$= -0.1455$$ and $$1.1455$$ Then inserting this in $f(x)$ $$f(-0.1455) = -0.01545$$ $$f(1.1455) = 0.2716$$
But looking at the graph it doesn't look like inflection points, so is it wrong or can any body help me?
Note that
$$f'(x) = 4x^3 - 6x^2 +2x\implies x=0,\, x=\frac12,\,x=1$$
$$f''(x) = 12x^2-12x+2=0 \implies x=\frac12\pm\frac{\sqrt 3}{6}$$
thus there are two inflection points.