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

Answer 1

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 role is

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

Do the first and second derivative calculations.

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

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

The turning points occur 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 time intervals to take into account are

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

Let's calculate the concavity by creating a variation chart.

#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 [^8.01, 8.01, -16.02, 16.01]}

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Answer 2

Paisley Johnson

To find points of inflection and determine the intervals of concavity for the function $y = x^3 - 2x^2 - 2$ , follow these steps:

Find the second derivative of the function.
Set the second derivative equal to zero and solve for $x$ to find potential points of inflection.
Determine the sign of the second derivative in the intervals between the potential points of inflection to determine the intervals of concavity.

Would you like me to explain any of these steps further?

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.