# What type of non-constant function has the same average rate of change and instantaneous rate of change over all intervals and for all values of "x?" Explain.

May 31, 2016

Any linear function $f \left(x\right) = A x + B$ satisfies this condition.
Any function that satisfies this condition is linear.

#### Explanation:

Obviously, a linear function $f \left(x\right) = A x + B$ satisfies this condition. Its average rate of change on any interval $\left[{x}_{1} , {x}_{2}\right]$ is equal to
$R = \frac{f \left({x}_{2}\right) - f \left({x}_{1}\right)}{{x}_{2} - {x}_{1}} = \frac{A {x}_{2} + B - A {x}_{1} - B}{{x}_{2} - {x}_{1}} = A$

Since this is true for any interval, instantaneous rate of change at any point ${x}_{1}$ is also equal to $A$ since it is a limit of average rate of change when the right end of an interval ${x}_{2}$ gets infinitely close to its left end ${x}_{1}$.

A little more interesting is to prove that this class of linear functions is the only set of functions defined for all real numbers having a property of the same rate of change on any interval as well as an instantaneous rate of change.

Here is a proof.
Let's fix two points on the X-axis: ${x}_{1}$, ${x}_{2}$ and take any other point $x$.
Since the average rate of change is the same on any interval,
$R = \frac{f \left({x}_{2}\right) - f \left({x}_{1}\right)}{{x}_{2} - {x}_{1}} = \frac{f \left(x\right) - f \left({x}_{1}\right)}{x - {x}_{1}}$

From the above we can conclude:
$R \left(x - {x}_{1}\right) = f \left(x\right) - f \left({x}_{1}\right)$
$f \left(x\right) = R x - R {x}_{1} + f \left({x}_{1}\right)$
As we see, $f \left(x\right)$ is a linear function, which is exactly what we wanted to prove.