How do you find all points of inflection given $y = x^{3} - 2 x^{2} + 1$ $y=x^3-2x^2+1$ ?

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Answer 1 · 2017-04-17T22:25:31

$(\frac{2}{3}, 0.41)$ $(2/3,0.41)$

Inflection points occur when the second derivative is equal to $0$ $0$

$\frac{d y}{d x} = 3 x^{2} - 4 x$ $dy/dx=3x^2-4x$

$\frac{d^{2} y}{{d x}^{2}} = 6 x - 4$ $(d^2y)/(dx^2)=6x-4$

Let $\frac{d^{2} y}{{d x}^{2}} = 0$ $(d^2y)/(dx^2)=0$

$0 = 6 x - 4$ $0=6x-4$

$6 x = 4$ $6x=4$

$x = \frac{4}{6} = \frac{2}{3}$ $x=4/6=2/3$

Solve for y-cordinate,

$y = {(\frac{2}{3})}^{3} - 2 {(\frac{2}{3})}^{2} + 1$ $y=(2/3)^3-2(2/3)^2+1$

$y = \frac{8}{27} - 2 (\frac{4}{9}) + 1$ $y=8/27-2(4/9)+1$

$y = \frac{8}{27} - \frac{8}{9} + 1$ $y=8/27-8/9+1$

$y = \frac{11}{27}$ $y=11/27$ or $0.41$ $0.41$

Therefore the point of inflection for the function $y = x^{3} - 2 x^{2} + 1$ $y=x^3-2x^2+1$ is $(\frac{2}{3}, 0.41)$ $(2/3,0.41)$

How do you find all points of inflection given y=x3−2x2+1y=x^3-2x^2+1?