How is the shape of the curve $f(x)$, near $x=a$, affected by $f'''(a)$ In introductory calculus classes we learn the utility of calculating $f'(x)$ and $f''(x)$ for sketching the curve $f(x)$. My question is given $f'''(a)$, how does this value affect the shape a curve $f(x)$ for $x$ near $a$?
Consider an example where $f'(x)>0$ and $f''(x)<0$ on the whole domain. How does the sign of $f'''(x)$ change the shape of the curve?
 A: Let's change the notation a little, to make things easier to read. Instead of a function $x \mapsto f(x)$, let's consider a function $t \mapsto x(t)$, and let's use dots to denote differentiation with respect to $t$.
Then the curvature of $x$ is given by 
$$
\kappa(t) = \frac{\ddot x}{\sqrt{1 + {\dot x}^2}}
$$
and the derivative of curvature is
$$
\dot\kappa = \frac{d\kappa}{dt} = \frac{\dddot x(1+ {\dot x}^2 ) - \dot x \ddot x^2  } { ({1 + {\dot x}^2})^{3/2}  }
$$
So, we can say the following:


*

*If $\dddot x > 0$ and $\dot x < 0$, then $\dot\kappa > 0$, so $\kappa$ is increasing, so radius of curvature is decreasing

*If $\dddot x < 0$ and $\dot x > 0$, then $\dot\kappa < 0$, so $\kappa$ is decreasing, so radius of curvature is increasing


So, the sign of $\dddot x$ gives us some information about growth of curvature in some situations, at least. 
But, in the unfortunate regions where $\dddot x > 0$ and $\dot x > 0$, or where $\dddot x < 0$ and $\dot x < 0$, this line of reasoning doesn't tell us anything about curvature, as far as I can see.
So, it's not quite as simple as I indicated in my original (lazy) answer. Thanks to @TedShifrin for pointing out the error in my reasoning.
A: You can also say that the third derivative $f′′′(x)$ gives you information about convexity (or concavity) of the first derivative (and geometric interpretation of the first derivative is obvious).
Maybe you have heard about the famous quotation of Richard Nixon: "The rate of increase of inflation is decreasing." That is a use of the third derivation in practice.
