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

Question

4582 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2017-04-17T22:25:31

$(2/3,0.41)$

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

$dy/dx=3x^2-4x$

$(d^2y)/(dx^2)=6x-4$

Let $(d^2y)/(dx^2)=0$

$0=6x-4$

$6x=4$

$x=4/6=2/3$

Solve for y-cordinate,

$y=(2/3)^3-2(2/3)^2+1$

$y=8/27-2(4/9)+1$

$y=8/27-8/9+1$

$y=11/27$ or $0.41$

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

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