Is there a way to determine the point with the highest instantaneous rate of change given a graph and a function? I'm trying to understand how to determine the greatest instantaneous rate of change for a given function. For example, for the function $f(x) = (0.5x^2 + x) - 0.05x^3$ is there a way to find and prove where the rate of change is highest? I can find the rate of change for any given value of $x$, but I don't see how to find the highest short of calculating it for each value of x and comparing them. The graph is x/y -- time/distance, so I can also see where the graph is steepest, but I'm curious to see if there is a way to be more precise than that. 
Thank you!
 A: You need to find the second derivative.
The candidates for the highest rate of change are among the points where the second derivative is either zero or it does not exist.
What you really want to do in to find the maximum value of the first derivative. 
In your case your function is a polynomial and the second derivative exists at every point.
All you have to do is to find zeros of the second derivative and compare the value of your first derivative at those points.        
A: If $f(x)=0.5x^2+x-0.05x^3$, then
$$f'(x)=x+1-0.15x^2$$
We are looking to maximise $f'(x)$, so we will take the second derivative and equate it to $0$.
\begin{align}
f''(x) & = 1-0.3x \\
0 & = 1-0.3x \\
0.3x & = 1 \\
x & = \frac{10}{3}
\end{align}
We need to check whether this is a maximum or minimum using the next derivative.
\begin{align}
f'''(x) & = -0.3 \\
f'''(\frac{10}{3}) & = -0.3
\end{align}
$f'''(\frac{10}{3}) \lt 0 \implies x=\frac{10}{3}$ is a maximum slope. 
A: Yes. As you may know, we find the maximum and minimum values of a function by calculating where the derivative is zero. So we can do this for the rate of change. For simplicity's sake, we'll let $v(x) = f'(x)$ be the rate of change of $f(x)$. To calculate the maximum of $v(x)$, we'll simply calculate $v'(x)$ and find zeros.
$$\begin{aligned}v(x) &= f'(x) \\
&=x+1-0.15x^2\end{aligned}$$
Then,
$$v'(x) = 1-0.3x$$
Solving for $v'(x)=0$, we find that $x=\frac 1{0.3}=\frac{10}3$. The graph will be sufficient to tell you that it is a maximum.
Therefore, the rate of change of $f(x)$ is highest when $x=10/3$. Additionally, we can find the rate of change of $f(x)$ here by finding $v(10/3)=6$.
