Find the formula for the a-unit moving average of a general exponential function. $f ( x ) = K e ^ { p x }$
A) $\bar { f } ( x ) = \frac { K } { p a } \left( e ^ { p x } + e ^ { p ( x - a ) } \right)$
B) $\bar { f } ( x ) = \frac { K } { a } \left( e ^ { p x } - e ^ { p ( x - p ) } \right)$
C) $\bar { f } ( x ) = \frac { K } { p a } \left( e ^ { p x } - e ^ { p ( x + a ) } \right)$
D) $\bar { f } ( x ) = \frac { K } { p } \left( e ^ { p x } - e ^ { a } \right)$
E) $\bar { f } ( x ) = \frac { K } { p a } \left( e ^ { p x } - e ^ { p ( x - a ) } \right)$

Question

Find the formula for the a-unit moving average of a general exponential function.   $f ( x ) = K e ^ { p x }$  
A) $\bar { f } ( x ) = \frac { K } { p a } \left( e ^ { p x } + e ^ { p ( x - a ) } \right)$
B) $\bar { f } ( x ) = \frac { K } { a } \left( e ^ { p x } - e ^ { p ( x - p ) } \right)$
C) $\bar { f } ( x ) = \frac { K } { p a } \left( e ^ { p x } - e ^ { p ( x + a ) } \right)$
D)   $\bar { f } ( x ) = \frac { K } { p } \left( e ^ { p x } - e ^ { a } \right)$
E) $\bar { f } ( x ) = \frac { K } { p a } \left( e ^ { p x } - e ^ { p ( x - a ) } \right)$

Quizplus · Accepted Answer

The correct formula for the a-unit moving average of a general exponential function $f(x) = Ke^{px}$ is derived by integrating the function over the interval from $x-a$ to $x$ and then dividing by the length of the interval, which is $a$. This results in $\bar{f}(x) = \frac{K}{pa}\left(e^{px} - e^{p(x-a)}\right)$, matching option E.

Find the Formula for the A-Unit Moving Average of a General