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Q35E

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Found in: Page 20

Linear Algebra With Applications

Book edition 5th
Author(s) Otto Bretscher
Pages 442 pages
ISBN 9780321796974

Find the polynomial f(t) of degree 3 such that ${\mathbit{f}}\left(1\right){\mathbf{=}}\left(1\right){\mathbf{,}}{\mathbit{f}}{\mathbf{\left(}}{\mathbf{2}}{\mathbf{\right)}}{\mathbf{=}}{\mathbf{5}}{\mathbf{,}}{\mathbit{f}}{\mathbf{\text{'}}}{\mathbf{\left(}}{\mathbf{1}}{\mathbf{\right)}}{\mathbf{=}}{\mathbf{2}}{\mathbf{,}}$and ${\mathbit{f}}{\mathbf{\text{'}}}\left(2\right){\mathbf{=}}{\mathbf{9}}$ , where ${\mathbit{f}}{\mathbf{\text{'}}}{\mathbf{\left(}}{\mathbit{t}}{\mathbf{\right)}}$ is the derivative of ${\mathbit{f}}\left(t\right)$ . Graph this polynomial.

The graphical representation of the cubic polynomial $f\text{'}\left(t\right)=-5+13t-10{t}^{2}+3{t}^{2}$ is,

See the step by step solution

Step 1: Consider the points and form the equations

The equation is, $f\left(t\right)=a+bt+c{t}^{2}+d{t}^{3}$

The derivative of the equation is, $f\text{'}\left(t\right)=b+2ct+3d{t}^{2}$

Substitute the points in the above equation to form the equations.

localid="1659349415086" $\left(1,1\right):\phantom{\rule{0ex}{0ex}}f\left(1\right)=a+b\left(1\right)+c{\left(1\right)}^{2}+d{\left(1\right)}^{3}\phantom{\rule{0ex}{0ex}}=1\phantom{\rule{0ex}{0ex}}a+b+c+d=1....\left(1\right)\phantom{\rule{0ex}{0ex}}\left(2,5\right):\phantom{\rule{0ex}{0ex}}f\left(2\right)=a+b\left(2\right)+c{\left(2\right)}^{2}+d{\left(2\right)}^{3}\phantom{\rule{0ex}{0ex}}=-1\phantom{\rule{0ex}{0ex}}a+2b+4c+8d=5......\left(2\right)\phantom{\rule{0ex}{0ex}}\left(1,2\right):\phantom{\rule{0ex}{0ex}}f\text{'}\left(1\right)=b\left(3\right)+2\left(c\right)+3d{\left(3\right)}^{2}\phantom{\rule{0ex}{0ex}}=2\phantom{\rule{0ex}{0ex}}b+2c+3d=2.......\left(3\right)\phantom{\rule{0ex}{0ex}}\left(2,9\right):\phantom{\rule{0ex}{0ex}}f\text{'}\left(2\right)=b\left(3\right)+2c{\left(3\right)}^{2}+3d{\left(3\right)}^{2}\phantom{\rule{0ex}{0ex}}=9\phantom{\rule{0ex}{0ex}}b+4c+12d=9.............\left(4\right)$

Step 2: Consider the matrix

Re-write the equations in terms of a matrix.

$\left|\begin{array}{c}a+b+c+d=1\\ a+2b+4c+8d=5\\ b+2c+3d=2\\ b+4c+12d=9\end{array}\right|$

The matrix form is,

$\left|\begin{array}{ccccc}1& 1& 1& 1& 1\\ 1& 2& 4& 8& 5\\ 0& 1& 2& 3& 2\\ 0& 1& 4& 12& 9\end{array}\right|$

Step 3: Solve the matrix

Consider the matrix.

$\left|\begin{array}{ccccc}1& 1& 1& 1& 1\\ 1& 2& 4& 8& 5\\ 0& 1& 2& 3& 2\\ 0& 1& 4& 12& 9\end{array}\right|$

Using row Echelon form to reduce the matrix.

$\left[\begin{array}{ccccc}1& 1& 1& 1& 1\\ 1& 2& 4& 8& 5\\ 0& 1& 2& 3& 2\\ 0& 1& 4& 12& 9\end{array}\right]=\left[\begin{array}{ccccc}1& 1& 1& 1& 1\\ 0& 1& 3& 7& 4\\ 0& 0& -1& -4& -2\\ 0& 0& 0& 1& 3\end{array}\right]\phantom{\rule{0ex}{0ex}}=\left[\begin{array}{ccccc}1& 0& 0& 0& -5\\ 0& 1& 0& 0& 13\\ 0& 0& 1& 0& -10\\ 0& 0& 0& 1& 3\end{array}\right]$

The values are, $a=-5,b=13,c=-10,d=3$

Therefore, the polynomial is,

$f\left(t\right)=a+bt+c{t}^{2}+d{t}^{3}\phantom{\rule{0ex}{0ex}}=-5+13t-10{t}^{2}+3{t}^{3}$

Step 4: Sketch the graph

The graphical representation of $f\left(t\right)=-5+13t-10{t}^{2}+3{t}^{3}$ is, ss