Use the appropriate Lagrange interpolating polynomials to find the cubic polynomial whose graph passes through the given points. (2, 1), (3, 1), (4, −3), (5, 0).

Question
Polynomial graphs
asked 2020-10-28
Use the appropriate Lagrange interpolating polynomials to find the cubic polynomial whose graph passes through the given points. \(\displaystyle{\left({2},\ {1}\right)},\ {\left({3},\ {1}\right)},\ {\left({4},\ −{3}\right)},\ {\left({5},\ {0}\right)}\).

Answers (1)

2020-10-29
Step 1 With \(\displaystyle{x}_{{{1}}}={2},\ {x}_{{{2}}}={3},\ \text{and}\ {x}_{{{4}}}={5}\), the Langrange interpolating polynomials give: \(\displaystyle{p}_{{{1}}}{\left({x}\right)}=\ {\frac{{{\left({x}\ -\ {x}_{{{2}}}\right)}{\left\lbrace{x}\ -\ {x}_{{{3}}}\right)}{\left({x}\ -\ {x}_{{{4}}}\right)}}}{{{\left({x}_{{{1}}}\ -\ {x}_{{{2}}}\right)}{\left({x}_{{{1}}}\ -\ {x}_{{{3}}}\right)}{\left({x}_{{{1}}}\ -\ {x}_{{{4}}}\right)}}}}\)
\(\displaystyle=\ {\frac{{{\left({x}\ -\ {3}\right)}{\left({x}\ -\ {4}\right)}{\left({x}\ -\ {5}\right)}}}{{{\left({2}\ -\ {3}\right)}{\left({2}\ -\ {4}\right)}{\left({2}\ -\ {5}\right)}}}}\)
\(\displaystyle=\ {\frac{{{\left({x}^{{{2}}}\ -\ {7}{x}\ +\ {12}\right)}{\left({x}\ -\ {5}\right)}}}{{{\left(-{1}\right)}{\left(-{2}\right)}{\left(-{3}\right)}}}}\)
\(\displaystyle=\ -\ {\frac{{{x}^{{{3}}}\ -\ {5}{x}^{{{2}}}\ -\ {7}{x}^{{{2}}}\ +\ {35}{x}\ +\ {12}{x}\ -\ {60}}}{{{6}}}}\)
\(\displaystyle{\color{red}{=\ -\ {\frac{{{1}}}{{{6}}}}{x}^{{{3}}}\ +\ {2}{x}^{{{2}}}\ -\ {\frac{{{47}}}{{{6}}}}{x}\ +\ {10}}}\)
\(\displaystyle{p}_{{{2}}}{\left({x}\right)}=\ {\frac{{{\left({x}\ -\ {x}_{{{1}}}\right)}{\left({x}\ -\ {x}_{{{3}}}\right)}{\left({x}\ -\ {x}_{{{4}}}\right)}}}{{{\left({x}_{{{2}}}\ -\ {x}_{{{1}}}\right)}{\left({x}_{{{2}}}\ -\ {x}_{{{3}}}\right)}{\left({x}_{{{2}}}\ -\ {x}_{{{4}}}\right)}}}}\)
\(\displaystyle=\ {\frac{{{\left({x}\ -\ {2}\right)}{\left({x}\ -\ {4}\right)}{\left({x}\ -\ {5}\right)}}}{{{\left({3}\ -\ {2}\right)}{\left({3}\ -\ {4}\right)}{\left({3}\ -\ {5}\right)}}}}\)
\(\displaystyle=\ {\frac{{{\left({x}^{{{2}}}\ -\ {6}{x}\ +\ {8}\right)}{\left({x}\ -\ {5}\right)}}}{{{\left({1}{\left(-{1}\right)}{\left(-{2}\right)}\right)}}}}\)
\(\displaystyle=\ {\frac{{{x}^{{{3}}}\ -\ {11}{x}^{{{2}}}\ +\ {38}{x}\ -\ {40}}}{{{2}}}}\)
\(\displaystyle{\color{red}{=\ {\frac{{{1}}}{{{6}}}}{x}^{{{3}}}\ -\ {\frac{{{11}}}{{{2}}}}{x}^{{{2}}}\ +\ {19}{x}\ -\ {20}}}\) Step 2 \(\displaystyle{p}_{{{3}}}{\left({x}\right)}=\ {\frac{{{\left({x}\ -\ {x}_{{{1}}}\right)}{\left({x}\ -\ {x}_{{{2}}}\right)}{\left({x}\ -\ {x}_{{{4}}}\right)}}}{{{\left({x}_{{{3}}}\ -\ {x}_{{{1}}}\right)}{\left({x}_{{{3}}}\ -\ {x}_{{{2}}}\right)}{\left({x}_{{{3}}}\ -\ {x}_{{{4}}}\right)}}}}\)
\(\displaystyle=\ {\frac{{{\left({x}\ -\ {2}\right)}{\left({x}\ -\ {3}\right)}{\left({x}\ -\ {5}\right)}}}{{{\left({4}\ -\ {2}\right)}{\left({4}\ -\ {3}\right)}{\left({4}\ -\ {5}\right)}}}}\)
\(\displaystyle=\ {\frac{{{\left({x}^{{{2}}}\ -\ {5}{x}\ +\ {6}\right)}{\left({x}\ -\ {5}\right)}}}{{{2}\ \cdot\ {1}{\left(-{1}\right)}}}}\)
\(\displaystyle=\ {\frac{{{x}^{{{3}}}\ -\ {10}{x}^{{{2}}}\ +\ {31}{x}\ -\ {30}}}{{{2}}}}\)
\(\displaystyle{\color{red}{=\ {\frac{{{1}}}{{{2}}}}{x}^{{{3}}}\ +\ {5}{x}^{{{2}}}\ -\ {\frac{{{31}}}{{{2}}}}{x}\ +\ {15}}}\)
\(\displaystyle{p}_{{{4}}}{\left({x}\right)}=\ {\frac{{{\left({x}\ -\ {x}_{{{1}}}\right)}{\left({x}\ -\ {x}_{{{2}}}\right)}{\left({x}\ -\ {x}_{{{3}}}\right)}}}{{{\left({x}_{{{4}}}\ -\ {x}_{{{1}}}\right)}{\left({x}_{{{4}}}\ -\ {x}_{{{2}}}\right)}{\left({x}_{{{4}}}\ -\ {x}_{{{3}}}\right)}}}}\)
\(\displaystyle=\ {\frac{{{\left({x}\ -\ {2}\right)}{\left({x}\ -\ {3}\right)}{\left({x}\ -\ {4}\right)}}}{{{\left({5}\ -\ {2}\right)}{\left({5}\ -\ {3}\right)}{\left({5}\ -\ {4}\right)}}}}\)
\(\displaystyle=\ {\frac{{{\left({x}^{{{2}}}\ -\ {5}{x}\ +\ {6}\right)}{\left({x}\ -\ {4}\right)}}}{{{3}\ \cdot\ {2}\ \cdot\ {1}}}}\)
\(\displaystyle=\ {\frac{{{x}^{{{3}}}\ -\ {9}{x}^{{{2}}}\ +\ {26}{x}\ -\ {24}}}{{{6}}}}\)
\(\displaystyle{\color{red}{=\ {\frac{{{1}}}{{{6}}}}{x}^{{{3}}}\ -\ {\frac{{{3}}}{{{2}}}}{x}^{{{2}}}\ +\ {\frac{{{13}}}{{{3}}}}{x}\ -\ {4}}}\) Step 3 With \(\displaystyle{y}_{{{1}}}={1},\ {y}_{{{2}}}=\ -{1},\ {y}_{{{3}}}={3}\ {\quad\text{and}\quad}\ {y}_{{{4}}}={0},\) Formula (9) gives: \(\displaystyle{p}{\left({x}\right)}={1}{\left(-\ {\frac{{{1}}}{{{6}}}}{x}^{{{3}}}\ +\ {2}{x}^{{{2}}}\ -\ {\frac{{{47}}}{{{6}}}}{x}\ +\ {10}\right)}-\ {1}{\left({\frac{{{1}}}{{{2}}}}{x}^{{{3}}}\ -\ {\frac{{{11}}}{{{2}}}}{x}^{{{2}}}\ +\ {19}{x}\ -\ {20}\right)}\)
\(\displaystyle+\ {3}{\left(-\ {\frac{{{1}}}{{{2}}}}{x}^{{{3}}}\ +\ {5}{x}^{{{2}}}\ -\ {\frac{{{31}}}{{{2}}}}{x}\ +\ {15}\right)}+\ {0}{\left({\frac{{{1}}}{{{6}}}}{x}^{{{3}}}\ -\ {\frac{{{3}}}{{{2}}}}{x}^{{{2}}}\ +\ {\frac{{{13}}}{{{3}}}}{x}\ -\ {4}\right)}\)
\(\displaystyle={x}^{{{3}}}{\left(-\ {\frac{{{1}}}{{{6}}}}\ -\ {\frac{{{1}}}{{{2}}}}\ -\ {\frac{{{3}}}{{{2}}}}\right)}\ +\ {x}^{{{2}}}{\left({2}\ +\ {\frac{{{11}}}{{{2}}}}\ +\ {15}\right)}\ +\ {x}{\left(-\ {\frac{{{47}}}{{{6}}}}\ -\ {19}\ -\ {\frac{{{92}}}{{{2}}}}\right)}\ +\ {10}\ -\ {20}\ +\ {45}\)
\(\displaystyle{\color{red}{=\ {\frac{{{11}}}{{{2}}}}{x}^{{{3}}}\ -\ {\frac{{{37}}}{{{2}}}}{x}^{{{2}}}\ +\ {\frac{{{173}}}{{{3}}}}{x}\ -\ {55}}}\)
0

Relevant Questions

asked 2020-11-09
Use the appropriate Lagrange interpolating polynomials to find the cubic polynomial whose graph passes through the given points. \(\displaystyle{\left({1},{2}\right)},{\left({2},{1}\right)},{\left({3},{3}\right)},{\left({6},{1}\right)}{\left({1},{2}\right)},{\left({2},{1}\right)},{\left({3},{3}\right)},{\left({6},{1}\right)}\).
asked 2021-02-14
Find the cubic polynomial whose graph passes through the points \(\displaystyle{\left(-{1},-{1}\right)},{\left({0},{1}\right)},{\left({1},{3}\right)},{\left({4},-{1}\right)}.\)
asked 2021-02-11
Find the cubic polynomial whose graph passes through the points \(\displaystyle{\left(-{1},-{1}\right)},{\left({0},{1}\right)},{\left({1},{3}\right)},{\left({4},-{1}\right)}.\)
asked 2021-03-08
For the following exercises, use the given information about the polynomial graph to write the equation. Double zero at \(\displaystyle{x}=−{3}\) and triple zero \(\displaystyle{a}{t}{x}={0}\). Passes through the point (1, 32).
asked 2021-02-20
Graph the polynomial in the given viewing rectangle. Find the coordinates of all local extrema. State each answer rounded to two decimal places. State the domain and range. \(\displaystyle{y}={x}^{{{3}}}-{3}{x}^{{{2}}},{\left[-{2},{5}\right]}{b}{y}{\left[-{10},{10}\right]}\)
asked 2020-12-06
For each polynomial function, one zero is given. Find all rational zeros and factor the polynomial. Then graph the function. f(x)=3x^{3}+x^{2}-10x-8ZSK, zero:2
asked 2021-01-06
Use your knowledge of the graphs of polynomial functions to make a rough sketch of the graph of \(\displaystyle{y}=-{2}{x}^{{{3}}}+{x}^{{{2}}}-{5}{x}+{2}\)
asked 2021-01-05
Let \(\displaystyle{S}_{{{N}}}{\left({x}\right)}={\frac{{{4}}}{{\pi}}}\ {\sum_{{{n}={1}}}^{{{N}}}}\ {\frac{{{1}\ -\ {\left(-{1}\right)}^{{{n}}}}}{{{n}^{{{3}}}}}}\ {\sin{{\left({n}{x}\right)}}}.\)
Construct graphs of \(\displaystyle{S}_{{{N}}}{\left({x}\right)}\ {\quad\text{and}\quad}\ {x}{\left(\pi\ -\ {x}\right)},\ {f}{\quad\text{or}\quad}\ {0}\ \leq\ {x}\ \leq\ \pi,\ {f}{\quad\text{or}\quad}\ {N}={2}\ {\quad\text{and}\quad}\ {t}{h}{e}{n}\ {N}={10}.\)
This will give some sense of the correctness of Fourier’s claim that this polynomial could be exactly represented by the infinite series
\(\displaystyle{\frac{{{4}}}{{\pi}}}\ {\sum_{{{n}={1}}}^{{\infty}}}\ {\frac{{{1}\ -\ {\left(-{1}\right)}^{{{n}}}}}{{{n}^{{{3}}}}}}\ {\sin{{\left({n}{x}\right)}}}\ {o}{n}\ {\left[{0},\ \pi\right]}.\)
asked 2021-01-31
The quadratic function \(\displaystyle{d}{i}{s}{p}{l}{a}{y}{s}{t}{y}\le{\left\lbrace{y}\right\rbrace}={\left\lbrace{a}\right\rbrace}{\left\lbrace{x}\right\rbrace}^{{{2}}}+{\left\lbrace{b}\right\rbrace}{\left\lbrace{x}\right\rbrace}+{\left\lbrace{c}\right\rbrace}\) whose graph passes through the points (1, 4), (2, 1) and (3, 4).
asked 2021-02-25
Graph the polynomial function. \(\displaystyle{f{{\left({x}\right)}}}=−{x}^{{{4}}}+{3}{x}^{{{3}}}−{x}+{1}\)
...