$\begingroup$I was doing very well in Calculus up until this point. I realize that concavity and $f'$ and $f''$ require one to really visualize what is happening with a function, but can someone please help me to understand this problem?
In particular, the second part of the question ask this:
So, if $f'(2) = 0$ and $f''(2) = 4$, what can you say about $f$?
What I know is this: if the $f'(2) = 0$, then $f$ has either a local maximum or minimum. $f''(2) = 4$ is greater than $0$, so I believe this means the graph is concave upward. So does this mean that the graph has a local minimum? If yes, is there any easier way to visualize these problems so that I can become more effective at solving them?
$\endgroup$31 Answer
$\begingroup$You're right, it has a local minimum; and I'm not sure it's possible to give an easier way of getting the answer than the one and a half lines it took you!
One thing you might say is that this is all about "what $f(x)$ looks like near $x=2$", and that the answer is that you're looking for the parabola which most closely fits with $f$ near this point.
This is a part of Taylor's thorem, which tells us here that $$f(x) \approx f(2) + (x-2) f'(2) + \frac{1}{2}(x-2)^2 f''(2) + \cdots = f(2) + 2(x-2)^2 +\cdots $$ where $(x-2)$ is small.
Another is to explain in detail that:
- $f'(2) = 0$ means that the graph is roughly flat at $2$.
- $f''(2) > 0$ means that $f' > 0$ to the right of $2$ but $f' < 0$ to the left.
- Therefore the graph comes down to this point from the left, and increases away from it to the right.
- Therefore the graph has a \_/ shape.
Edit: In response to your comment about $f''$, my usual way of thinking about it is this: it is the rate of change of $f'$. Hence $f''>0$ tells us the function's slope is gradually bending upwards. This means, since it's flat where $f'(2)=0$, it must be curved in a smile shape.
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