Find the quadratic polynomial g(x)-ax^{2} + bx + c text{which best fits the function} f(x)=e^{x} text{at} x=0, text{in the sense that} g(0)=f(0), text{and} g'(0)=f'(0), text{and} g''(0)=f''(0). Using a computer or calculator, sketch graphs of f and g on the same axes. What do you notice?

Question
Polynomial graphs
asked 2021-01-30
Find the quadratic polynomial \(\displaystyle{g{{\left({x}\right)}}}-{a}{x}^{{{2}}}\ +\ {b}{x}\ +\ {c}\ \text{which best fits the function}\ {f{{\left({x}\right)}}}={e}^{{{x}}}\ \text{at}\ {x}={0},\ \text{in the sense that}\ {g{{\left({0}\right)}}}={f{{\left({0}\right)}}},\ \text{and}\ {g}'{\left({0}\right)}={f}'{\left({0}\right)},\ \text{and}\ {g}{''}{\left({0}\right)}={f}{''}{\left({0}\right)}.\) Using a computer or calculator, sketch graphs of f and g on the same axes. What do you notice?

Answers (1)

2021-01-31
Step 1 The functions f(x) and g(x) are: \(\displaystyle{f{{\left({x}\right)}}}={e}^{{{x}}}\)
\(\displaystyle{g{{\left({x}\right)}}}={a}{x}^{{{2}}}\ +\ {b}{x}\ +\ {c}\)
\(\displaystyle{f}'{\left({x}\right)}=\ {\frac{{{d}{\left({f{{\left({x}\right)}}}\right)}}}{{{\left.{d}{x}\right.}}}}=\ {\frac{{{d}{\left({e}^{{{x}}}\right)}}}{{{\left.{d}{x}\right.}}}}\) Using formula for derivative of exponential function: \(\displaystyle{f}'{\left({x}\right)}={e}^{{{x}}}\)
\(\displaystyle{f}{''}{\left({x}\right)}=\ {\frac{{{d}{\left({f},{\left({x}\right)}\right)}}}{{{\left.{d}{x}\right.}}}}=\ {\frac{{{d}{\left({e}^{{{x}}}\right)}}}{{{\left.{d}{x}\right.}}}}\) Using formula for derivative of exponential function: \(\displaystyle{f}{''}{\left({x}\right)}={e}^{{{x}}}\)
\(\displaystyle{g}'{\left({x}\right)}=\ {\frac{{{d}{\left({g{{\left({x}\right)}}}\right)}}}{{{\left.{d}{x}\right.}}}}=\ {\frac{{{d}{\left({a}{x}^{{{2}}}\ +\ {b}{x}\ +\ {c}\right)}}}{{{\left.{d}{x}\right.}}}}\) Using Theorem 3.2: \(\displaystyle{g}'{\left({x}\right)}=\ {\frac{{{d}{\left({a}{x}^{{{2}}}\right)}}}{{{\left.{d}{x}\right.}}}}\ +\ {\frac{{{d}{\left({b}{x}\right)}}}{{{\left.{d}{x}\right.}}}}\ +\ {\frac{{{d}{\left({c}\right)}}}{{{\left.{d}{x}\right.}}}}\) Using Theorem 3.1: \(\displaystyle{g}'{\left({x}\right)}={a}\ \times\ {\frac{{{d}{\left({x}^{{{2}}}\right)}}}{{{\left.{d}{x}\right.}}}}\ +\ {b}\ \times\ {\frac{{{d}{\left({x}\right)}}}{{{\left.{d}{x}\right.}}}}\ +\ {0}\) Step 2 Using Power Law: \(\displaystyle{g}'{\left({x}\right)}={a}\ \times\ {2}\ \times\ {x}^{{{2}\ -\ {1}}}\ +\ {b}\ \times\ {1}\ +\ {0}\)
\(\displaystyle{g}'{\left({x}\right)}={2}{a}{x}\ +\ {b}\)
\(\displaystyle{g}{''}{\left({x}\right)}=\ {\frac{{{d}{\left({g}'{\left({x}\right)}\right)}}}{{{\left.{d}{x}\right.}}}}=\ {\frac{{{d}{\left({2}{a}{x}\ +\ {b}\right)}}}{{{\left.{d}{x}\right.}}}}\) Using Theorem 3.2: \(\displaystyle{g}{''}{\left({x}\right)}=\ {\frac{{{d}{\left({2}{a}{x}\right)}}}{{{\left.{d}{x}\right.}}}}\ +\ {\frac{{{d}{\left({b}\right)}}}{{{\left.{d}{x}\right.}}}}\) Using Theorem 3.1: \(\displaystyle{g}{''}{\left({x}\right)}={2}{a}\ \times\ {\frac{{{d}{\left({x}\right)}}}{{{\left.{d}{x}\right.}}}}\ +\ {0}\)
\(\displaystyle{g}{''}{\left({x}\right)}={2}{a}\ \times\ {1}\)
\(\displaystyle{g}{''}{\left({x}\right)}={2}{a}\) Substituting \(\displaystyle{x}={0}\ \text{we get}:\)
\(\displaystyle{f{{\left({0}\right)}}}={f}'{\left({0}\right)}={f}{''}{\left({0}\right)}={1}\)
\(\displaystyle{g}{''}{\left({0}\right)}={2}{a}\)
\(\displaystyle{g}'{\left({0}\right)}={b}\)
\(\displaystyle{g{{\left({0}\right)}}}={c}\) Using \(\displaystyle{g{{\left({0}\right)}}}={f{{\left({0}\right)}}},\ {g}'{\left({0}\right)}={f}'{\left({0}\right)}\ \text{and}\ {g}{''}{\left({0}\right)}={f}{''}{\left({0}\right)},\ \text{we get}:\)
\(\displaystyle{2}{a}={1}\)
\(\displaystyle{a}=\ {\frac{{{1}}}{{{2}}}}\)
\(\displaystyle{b}={1}\)
\(\displaystyle{c}={1}\) Thus the quadratic polynomial g(x) that fits f(x) is: \(\displaystyle{g{{\left({x}\right)}}}={\left({\frac{{{x}^{{{2}}}}}{{{2}}}}\ +\ {x}\ +\ {1}\right)}\) Step 3 The two functions look very much similar near \(\displaystyle{x}={0}.\) Both functions increase for large positive values of x, but \(\displaystyle{e}^{{{x}}}\) increases quickly compared to quadratic function. For negative values of x, the quadratic goes to infinity while the exponential goes to 0. The exponential function is and increasing function while quadratic function is initially decreasing till \(\displaystyle{x}=\ -{1}\) and then increasing function. image
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