# How do you find the average rate of change of f(x)= x^3 + 1 over the interval (2,3) and (-1,1)?

Mar 2, 2018

$\text{1) average rate of change of" \ f(x) = x^3 + 1, "over} \setminus \left[2 , 3\right] \setminus q \quad \setminus = 19.$

$\text{2) average rate of change of" \ f(x) = x^3 + 1, "over} \setminus \left[- 1 , 1\right] = 1.$

#### Explanation:

$\text{[Note, the average rate of change is defined over a closed}$
$\text{interval, for example, over the interval" \ [ -3, 5 ]. \ \ "It is not}$
$\text{defined over an open interval, such as the interval" \ (-7, 10).}$
$\text{When looking at the definition of this, as below,}$
$\text{this necessity will be clear. (I'll assume you meant closed}$
$\text{intervals in your question.) ]}$

$\text{Recall the definition of the average rate of change of a}$
$\text{function, over a closed interval:}$

$\text{average rate of change of" \ \ f(x), \ "over the interval} \setminus \setminus \left[a , b\right] \setminus =$

$\setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad \frac{f \left(b\right) - f \left(a\right)}{b - a} .$

$\text{So, for our examples here:}$

$\text{1) average rate of change of" \ f(x) = x^3 + 1,}$
$\setminus q \quad \setminus q \quad \text{over the interval} \setminus \left[2 , 3\right] =$

$\setminus q \quad \setminus q \quad \setminus q \quad \frac{f \left(3\right) - f \left(2\right)}{3 - 2} \setminus = \setminus \frac{\left[{3}^{3} + 1\right] - \left[{2}^{3} + 1\right]}{3 - 2}$

$\setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad \setminus \quad \setminus = \setminus \frac{\left[28\right] - \left[9\right]}{1}$

$\setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad \setminus \quad \setminus = \setminus 28 - 9 \setminus = \setminus 19.$

$\text{2) average rate of change of} \setminus f \left(x\right) = {x}^{3} + 1 ,$
$\setminus q \quad \setminus q \quad \text{over the interval} \setminus \left[- 1 , 1\right] =$

$\setminus q \quad \setminus q \quad \setminus q \quad \frac{f \left(1\right) - f \left(- 1\right)}{1 - \left(- 1\right)} \setminus = \setminus \frac{\left[{1}^{3} + 1\right] - \left[{\left(- 1\right)}^{3} + 1\right]}{1 - \left(- 1\right)}$

$\setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad \setminus \quad \setminus = \setminus \frac{\left[2\right] - \left[- 1 + 1\right]}{1 + 1}$

$\setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad \setminus \quad \setminus = \setminus \frac{2 - 0}{2} \setminus = \setminus \frac{2}{2} \setminus = \setminus 1.$

$\text{So, summarizing:}$

$\text{1) average rate of change of" \ f(x) = x^3 + 1, "over} \setminus \left[2 , 3\right] \setminus q \quad \setminus = 19.$

$\text{2) average rate of change of" \ f(x) = x^3 + 1, "over} \setminus \left[- 1 , 1\right] = 1.$