How do you find points of inflection and determine the intervals of concavity given $y=x^3-2x^2-2$ ?

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Answer 1 · 2018-07-23T15:16:45

The function is concave when $x in (-oo,2/3)$ and convex when $(2/3,+oo)$ . The point of inflection is at $(0.667,-2.593)$

The function is

$f(x)=x^3-2x^2-2$

Calculate the first and second derivatives

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

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

The points of inflection are when

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

That is

$6x-4=0$

$=>$ , $x=4/6=2/3$

The point of inflection is at $(0.667,-2.593)$

The intervals to consider are

$I_1=(-oo,2/3)$ and $I_2=(2/3,+oo)$

Let's build a variation chart to determine the concavity

$color(white)(aaaa)$ $" Interval "$ $color(white)(aaaa)$ $(-oo,2/3)$ $color(white)(aaaa)$ $(2/3,oo)$

$color(white)(aaaa)$ $" Sign "(d^2y)/dx^2$ $color(white)(aaaaaa)$ $(-)$ $color(white)(aaaaaaaaa)$ $(+)$

$color(white)(aaaa)$ $" Concavity "$ $color(white)(aaaaaa)$ $nn$ $color(white)(aaaaaaaaaaa)$ $uu$

graph{x^3-2x^2-2 [-16.02, 16.01, -8.01, 8.01]}

How do you find points of inflection and determine the intervals of concavity given y=x^3-2x^2-2y=x^3-2x^2-2?