Question

find the limit as x approaches c of the average rate of change of each function from c to x.

$c=0;f\left(x\right)=3x^3-2x^2+4$

Short answer

The limit is$0$

Step-by-step solution

Step 1. Given information is 

$c=0;f\left(x\right)=3x^3-2x^2+4$

Step 2. An average rate change of function f  from point c to x

$$\begin{gathered} \frac{∆y}{∆x}=\frac{f\left(x\right)-f\left(c\right)}{x-c} \\ \frac{∆y}{∆x}=\frac{3x^3-2x^2+4-(3c^3-2c^2+4)}{x-c} \\ \frac{∆y}{∆x}=\frac{3x^3-2x^2+4-(3{\left(0\right)}^3-2{\left(0\right)}^2+4)}{x-0} \\ \frac{∆y}{∆x}=\frac{3x^3-2x^2+4-4}{x} \\ \frac{∆y}{∆x}=\frac{3x^3-2x^2}{x} \\ \frac{∆y}{∆x}=\frac{x(3x^2-2x)}{x} \\ \frac{∆y}{∆x}=3x^2-2x \end{gathered}$$

Step 3. Solving for limit we get,

$$\begin{gathered} \underset{x\to c}{\lim }\frac{∆y}{∆x}=\underset{x\to c}{\lim }\frac{f\left(x\right)-f\left(c\right)}{x-c} \\ \underset{x\to c}{\lim }\frac{∆y}{∆x}=\underset{x\to c}{\lim }(3x^2-2x) \\ \underset{x\to c}{\lim }\frac{∆y}{∆x}=3{\left(0\right)}^2-2\left(0\right) \\ \underset{x\to c}{\lim }\frac{∆y}{∆x}=0 \end{gathered}$$