Problem+23+Carolyn+and+Abby+V.

23. Suppose f'' is continuous on (-infinity, infinity).

a) If f'(2) = 0 and f(2) = -5 what can you say about f? If f'(2) = 0 then at x = 2 there is either a local minimum or local maximum, as this is where f(x) goes from increasing to decreasing (or vice versa). If f(2) = -5 then at x = 2 we can conclude that there is a local maximum because having a negative second derivative means that f'(x) is decreasing.and that f(x) is concave down. Based on the second derivative test, this makes f(2) a local maximum.

One comment: just because f ' (2) = 0 we are not guaranteed that we have a local max or min -- all we know is that we have a stationary point. It is the additional fact that f " (2) is negative that makes a difference, and allows us to conclude that it is not just a "resting" point.

b) If f'(6) = 0 and f(6) = 0 what can you say about f? If f'(6) = 0 then at x = 6 there is a critical point. This is not a local maximum or minimum because the graph is changing from concave down to concave up (or vice versa) as f(6) is zero and therefore cannot be at the top of a hill or the bottom of a valley. If f''(6) = 0 then at x = 6 there is an inflection point, meaning the f(x) graph is going from concave up to concave down (or vice versa) at x=6. It is possible that there is an inflection point at x = 6 -- however, we really can't be assured of this because we don't know if the second derivative changes sign at that point -- which is really our only guarantee of an inflection point! What we do know is that we don't have enough information to draw a conclusion based on the second derivative test!