Question

find the relative rate of change of (f(x)=110x - 0.2x^{2}). the relative rate of change of (f(x)) is (square).

Explanation:

Step1: Recall the formula for relative rate of change

The relative rate of change of a function $y = f(x)$ is given by $\frac{f'(x)}{f(x)}$. First, find the derivative of $f(x)=110x - 0.2x^{2}$.
Using the power - rule $\frac{d}{dx}(ax^{n})=nax^{n - 1}$, where $a$ is a constant and $n$ is a real number.
$f'(x)=\frac{d}{dx}(110x-0.2x^{2})=\frac{d}{dx}(110x)-\frac{d}{dx}(0.2x^{2})$.
$f'(x)=110 - 0.4x$.

Step2: Calculate the relative rate of change

The relative rate of change is $\frac{f'(x)}{f(x)}=\frac{110 - 0.4x}{110x-0.2x^{2}}$.

Answer:

$\frac{110 - 0.4x}{110x-0.2x^{2}}$