For y = f(x) = x^3, find the average rate of change of y with respect to x as x changes from 3 to 5.

Step-by-step solution

1. 1

   Write the definition. The average rate of change of $f$ over $[a,b]$ is $\frac{f(b)-f(a)}{b-a}$ — the slope of the secant line through $(a,f(a))$ and $(b,f(b))$, not the derivative at a single point.

2. 2

   Evaluate the endpoints. $f(5)=5^3=125$ and $f(3)=3^3=27$.

3. 3

   Form the change in $y$. $\Delta y=125-27=98$.

4. 4

   Form the change in $x$. $\Delta x=5-3=2$.

5. 5

   Divide. $\frac{\Delta y}{\Delta x}=\frac{98}{2}=49$, so on average $y$ rises $49$ units per unit increase in $x$ across this interval.

6. 6

   Compare with the instantaneous rates. Since $f'(x)=3x^2$, the slope is $27$ at $x=3$ and $75$ at $x=5$; the average $49$ lies between them, exactly as the Mean Value Theorem requires.