What are the points of inflection, if any, of #f(x)=2x^3 + 8x^2 #?

Answer 1

At #x=-4/3# there is point of inflection

Given -

#y=2x^3+8x^2#
To find the point of inflection put send derivative equal to zero and find the value of #x#
#dy/dx=6x^2+16x#
#(d^2y)/(dx^2)=12x+16#
#(d^2y)/(dx^2)=0 => 12x+16=0#
#x=(-16)/12=-4/3#

graph{[-10, 10, -5, 5]} = 2x^3 + 8x^2

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Answer 2

To find the points of inflection, we need to calculate the second derivative of the function, f(x)=2x^3 + 8x^2, and then solve for values of x where the second derivative equals zero or is undefined.

First derivative: f'(x) = 6x^2 + 16x Second derivative: f''(x) = 12x + 16

Setting f''(x) equal to zero and solving for x:

12x + 16 = 0 x = -16/12 x = -4/3

So, the point of inflection is at x = -4/3.

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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.

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.

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.

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.

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