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

Question
Confidence intervals
asked 2021-01-02
Given \(\displaystyle{f{{\left({x}\right)}}}={x}^{{{2}}}+{3}{x}+{8}\), find the average rate of change of \(\displaystyle{f{{\left({x}\right)}}}\) over each of the following pairs of intervals.

Answers (1)

2021-01-03
Step 1 Given function is \(\displaystyle{f{{\left({x}\right)}}}={x}^{{{2}}}+{3}{x}+{8}\) Firstly, we will find the value of \(\displaystyle{f{{\left({x}\right)}}}{a}{t}{x}={1.9},{1.99},{2},{2.1},{2.01}\) \(\displaystyle{f{{\left({1.9}\right)}}}={\left({1.9}\right)}^{{{2}}}+{3}{\left({1.9}\right)}+{8}={17.31}\)
\(\displaystyle{f{{\left({1.99}\right)}}}={\left({1.99}\right)}^{{{2}}}+{3}{\left({1.99}\right)}+{8}={17.9301}\)
\(\displaystyle{f{{\left({2}\right)}}}={\left({2}\right)}^{{{2}}}+{3}{\left({2}\right)}+{8}={18}\)
\(\displaystyle{f{{\left({2.01}\right)}}}={\left({2.01}\right)}^{{{2}}}+{3}{\left({2.01}\right)}+{8}={18.0701}\)
\(\displaystyle{f{{\left({2.1}\right)}}}={\left({2.1}\right)}^{{{2}}}+{3}{\left({2.1}\right)}+{8}={18.71}\) Step 2 Average rate of change of f(x) over [1.9, 2] is \(\displaystyle{A}_{{{1}}}={\frac{{{f{{\left({2}\right)}}}-{f{{\left({1.9}\right)}}}}}{{{2}-{1.9}}}}={\frac{{{18}-{17.31}}}{{{0.1}}}}={6.9}\) Average rate of change of f(x) over [1.99,2] is \(\displaystyle{A}_{{{2}}}={\frac{{{f{{\left({2}\right)}}}-{f{{\left({1.99}\right)}}}}}{{{2}-{1.99}}}}={\frac{{{18}-{17.9301}}}{{{0.01}}}}={6.99}\) Average rate of change of f(x) over [2,2.1] is \(\displaystyle{A}_{{{3}}}={\frac{{{f{{\left({2.1}\right)}}}-{f{{\left({2}\right)}}}}}{{{2.1}-{2}}}}={\frac{{{18.71}-{18}}}{{{0.1}}}}={7.1}\) Average rate of change of f(x) over [2,2.01] is \(\displaystyle{A}_{{{4}}}={\frac{{{f{{\left({2.01}\right)}}}-{f{{\left({2}\right)}}}}}{{{2.01}-{2}}}}={\frac{{{18.0701}-{18}}}{{{0.01}}}}={7.01}\) Step 3 (c) Instantaneous rate of change of f(x) at \(\displaystyle{x}={2}{i}{s}\) \(\displaystyle{f}'{\left({2}\right)}=\lim_{{{x}\rightarrow{2}}}{\frac{{{f{{\left({x}\right)}}}-{f{{\left({2}\right)}}}}}{{{x}-{2}}}}\) From above parts as x approaches 2, \(\displaystyle{\frac{{{f{{\left({x}\right)}}}-{f{{\left({2}\right)}}}}}{{{x}-{2}}}}\) approaches 7. Therefore, Instantaneous rate of change of \(\displaystyle{f{{\left({x}\right)}}}\) at \(\displaystyle{z}={2}{i}{s}{f}'{\left({2}\right)}={7}\) 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
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