Suppose that f′′′is continuous and f′(c) =f′′(c) = 0, but f′′′(c)>0. Does f(x) have a local maximum, minimum or a point of inflection at c?

Answer 1

Given: f(x) where c will be x in the proof, where c is on the horizontal line x-axis in a graph so we have f(c). f'(c) = f'(x) = slope. f''(x) = f'(c) = concavity and f''(c) = f''(x) shows whether concavity is increasing or decreasing. Since f''(x) = f''(c) = 0, we know that f(c) = f(x) does not have a local maximum or local minimum there. NOTE: F''(x) or F"(c)>0, means you have a local minimum, and f''(x) or f''(c)<0 means you have a local maximum, and f'(x) or f'(c) = 0, now knowing f'(x) or f'(c) = 0 means you have an inflection point. Additionally, one needs the lowest order non-zero derivative to be of odd order (1st, 3rd, 5th, etc.). If the lowest order non-zero derivative is of even order, the point is not a point of inflection.

Since f''(x) or f"(c)>0 we know concavity is increasing, so we can assume that at least one term in the function is odd, and the function is not symmetric, and does have an inflection (point) at f(c).

ANOTHER ANSWER: Suppose we have f(4) that is continuous, and f;(c) = f''(c) = f'''(c) = 0. Does f(x) have a local maximum, local minimum, or point of inflection at c? Justify the answer.

Given: f(4) = continuous and f(4) ≠ 0

Given: f'(c) = f""(c) = f''' = 0 and f(4) ≠ 0

f'(c) = 0, means c is a critical point to f(x).

From the Second Derivative Test if f"(c)>0, then f(x) has a local minimum at x = c.

From the Second Derivative Test if f"(x)<0, then f(x) has a local maximum at x = c.

However, given f"(c) = 0 in this problem, then we conclude that there is not any local minimum or local maximum.

We f"(c) = 0, and c = inflection point

C: We can define from the Second Derivative Test, f"(x) = 0, f(x) has an inflection point at x, and x = c, and we conclude f(x) has an inflection point at x = c.