# Given f(x) = x^{2} + 3x + 8, find the average rate of change of f(x) over each of the following pairs of intervals.

Given $f\left(x\right)={x}^{2}+3x+8$, find the average rate of change of $f\left(x\right)$ over each of the following pairs of intervals.
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Step 1
Given function is $f\left(x\right)={x}^{2}+3x+8$ Firstly, we will find the value of $f\left(x\right)atx=1.9,1.99,2,2.1,2.01$ $f\left(1.9\right)={\left(1.9\right)}^{2}+3\left(1.9\right)+8=17.31$
$f\left(1.99\right)={\left(1.99\right)}^{2}+3\left(1.99\right)+8=17.9301$
$f\left(2\right)={\left(2\right)}^{2}+3\left(2\right)+8=18$
$f\left(2.01\right)={\left(2.01\right)}^{2}+3\left(2.01\right)+8=18.0701$
$f\left(2.1\right)={\left(2.1\right)}^{2}+3\left(2.1\right)+8=18.71$
Step 2
Average rate of change of f(x) over [1.9, 2] is ${A}_{1}=\frac{f\left(2\right)-f\left(1.9\right)}{2-1.9}=\frac{18-17.31}{0.1}=6.9$ Average rate of change of f(x) over [1.99,2] is ${A}_{2}=\frac{f\left(2\right)-f\left(1.99\right)}{2-1.99}=\frac{18-17.9301}{0.01}=6.99$ Average rate of change of f(x) over [2,2.1] is ${A}_{3}=\frac{f\left(2.1\right)-f\left(2\right)}{2.1-2}=\frac{18.71-18}{0.1}=7.1$ Average rate of change of f(x) over [2,2.01] is ${A}_{4}=\frac{f\left(2.01\right)-f\left(2\right)}{2.01-2}=\frac{18.0701-18}{0.01}=7.01$
Step 3
(c) Instantaneous rate of change of f(x) at $x=2is$ ${f}^{\prime }\left(2\right)=\underset{x\to 2}{lim}\frac{f\left(x\right)-f\left(2\right)}{x-2}$ From above parts as x approaches 2, $\frac{f\left(x\right)-f\left(2\right)}{x-2}$ approaches 7. Therefore, Instantaneous rate of change of $f\left(x\right)$ at $z=2is{f}^{\prime }\left(2\right)=7$ Step 4 Ans: Average rate of change of f(x) over [1.9,2] is 6.9 Average rate of change of f(x) over [1.99,2] is 6.99 Average rate of change of f(x) over [2,2.1] is 7.1 Average rate of change of f(x) over [2,2.01] is 7.01 Instantaneous rate of change of f(x) at x=2 is 7