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

The function is concave when

The role is

Do the first and second derivative calculations.

The turning points occur when

That is

The time intervals to take into account are

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

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

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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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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.

- How do you use the second derivative test to find the local maximum and minimum for #f(x)=x^4-4x^3+4x^2+6#?
- Can an asymptote be an inflection point?
- How do you find local maximum value of f using the first and second derivative tests: #f(x) = 7e^x#?
- How does the first derivative test differ from the second derivative test?
- Is #f(x)=x^3-xe^(x-x^2) # concave or convex at #x=1 #?

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