If f(2)=3, f'(2)=1, and f''(2)=0, is f(2,3) an inflection point? Why or why not?

It possibly could be.  It all depends on whether the function is concave up on one side of 2 and concave down on the other side of 2. If that is the case, it is an inflection point.
From what is given, we don't know if it is concave up or concave down on either side.

It could be y = x+1, for then f(2) = 3, f'(x) = 1, so f'(2) = 1, and f"(x) = 0, so f"(2) = 0.
In this case, neither side if concave up or down.

If we took f"(x) = 2-x, that would make f"(2) = 2.
That would make f'(x) = 2x - (x^2)/2 + C.
Since f'(2) = 1 and we can see that f'(2) = 2 + C, C is -1.
This makes f'(x) = 2x - (x^2)/2 - 1.

From there f(x) = x^2 - (x^3)/6 - x + C.
Since f(2) = 3 and f(2) can be seen to be 4 -4/3 - 2 + C = 14/3 + C,
that means C = 1/3, so f(x) = x^2 - (x^3)/6 - x + 1/3.

As can be seen, this is concave up on the left and concave down on the right.  


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