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

Question

1517 views around the world

You can reuse this answer
Creative Commons License

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]}