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

Answer 1

Paisley Johnson

#(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)#

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Answer 2

Ezekiel Alexander

To find all points of inflection for the function y = x^3 - 2x^2 + 1, follow these steps:

Compute the second derivative of the function.
Set the second derivative equal to zero and solve for the values of x.
Substitute these values of x back into the original function to find the corresponding y-values.
The points (x, y) obtained from step 3 represent the points of inflection for the function.

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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.