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

Solve for y-cordinate,

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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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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.

- Is #f(x)=-2x^5-2x^3-12x^2-6x+1# concave or convex at #x=-1#?
- For what values of x is #f(x)= -5x^3+2x^2-3x-12 # concave or convex?
- What is the x-coordinate of the point of inflection on the graph of #y= 1/10x^5#?
- For what values of x is #f(x)=x^3-e^x# concave or convex?
- How do you find points of inflection and determine the intervals of concavity given #y=x^2-x-1#?

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