Question # Given the following function: f(x)=1.01e^{4x}-4.62e^{3x}-3.11e^{2x}+12.2e^{x} - 1.99 a)Use three-digit rounding frithmetic, the assumption that e^{1.5

Polynomial arithmetic
ANSWERED Given the following function: $$\displaystyle{f{{\left({x}\right)}}}={1.01}{e}^{{{4}{x}}}-{4.62}{e}^{{{3}{x}}}-{3.11}{e}^{{{2}{x}}}+{12.2}{e}^{{{x}}}-{1.99}$$ a)Use three-digit rounding frithmetic, the assumption that $$\displaystyle{e}^{{{1.53}}}={4.62}$$, and the fact that $$\displaystyle{e}^{{{n}{x}}}={\left({e}^{{{x}}}\right)}^{{{n}}}$$ to evaluate $$\displaystyle{f{{\left({1.53}\right)}}}$$ b)Redo the same calculation by first rewriting the equation using the polynomial factoring technique c)Calculate the percentage relative errors in both part a) and b) to the true result $$\displaystyle{f{{\left({1.53}\right)}}}=-{7.60787}$$ 2020-11-02

a) $$\displaystyle{f{{\left({x}\right)}}}={1.01}{e}^{{{4}{x}}}-{4.62}{e}^{{3}}{x}-{3.11}{e}^{{{2}{x}}}+{12.2}{e}^{{x}}-{1.99}$$

$${f{{\left({x}\right)}}}={1.01}{\left({e}^{{x}}\right)}^{{4}}-{4.62}{\left({e}^{{x}}\right)}^{{3}}-{3.11}{\left({e}^{{{x}}}\right)}^{{2}}-{12.2}{e}^{{x}}-{1.99}{\left({a}{s}\ {e}^{{{n}{x}}}={\left({e}^{{x}}\right)}^{{n}}\right)}$$

$${f{{\left({1.53}\right)}}}={1.01}{\left({e}^{{{1.53}}}\right)}^{{4}}-{4.62}{\left({e}^{{{1.53}}}\right)}^{{3}}-{3.11}{\left({e}^{{{1.53}}}\right)}^{{2}}+{12.2}{e}^{{{1.53}}}-{1.99}$$

$$={1.01}{\left({4.62}\right)}^{{4}}-{4.62}{\left({4.62}\right)}^{{3}}-{3.11}{\left({4.62}\right)}^{{2}}+{12.2}{\left({4.62}\right)}-{1.99}{\left({a}{s}\ {e}^{{{1.53}={4.62}}}\right)}$$

$$={1.01}{\left({455.583}\right)}-{4.62}{\left({98.611}\right)}-{3.11}{\left({21.344}\right)}+{56.364}-{1.99}$$

$$={460.139}-{455.583}-{66.380}+{54.374}$$

$$=-{7.45}$$

Therefore, the value of f(1.53) abtained by this method is -7.45.

b) The given function can be factorized sa follows. $$\displaystyle{f{{\left({x}\right)}}}={1.01}{\left({e}^{{x}}+{1.715}\right)}{\left({e}^{{x}}-{0.173}\right)}{\left({e}^{{x}}-{1.415}\right)}{\left({e}^{{x}}-{4.702}\right)}$$ On substituting $$\displaystyle{x}={1.53}$$ and using $$\displaystyle{e}^{{{1.53}}}={4.62},$$ we get $$\displaystyle{f{{\left({1.53}\right)}}}={1.01}{\left({4.62}+{1.715}\right)}{\left({4.62}-{0.173}\right)}{\left({4.62}-{1.415}\right)}{\left({4.62}-{4.702}\right)}$$

$$={1.01}{\left({6.335}\right)}{\left({4.447}\right)}{\left({3.205}\right)}{\left(-{0.082}\right)}$$

$$=-{7.478}$$

Therefore, the value of $$\displaystyle{f{{\left({1.53}\right)}}}$$ obtained by this method is -7.478.

c) Percentage error $$\displaystyle\delta$$ is given by $$\displaystyle\delta={\left|{\frac{{\nu_{{A}}-\nu_{{E}}}}{{\nu_{{E}}}}}\right|}{100}\%$$ Here $$\displaystyle\nu_{{A}}$$ is the actual value and $$\displaystyle\nu_{{E}}$$ is the expected value which is -7.60787 in this case. For the value obtained in part (a), the percentage error is $$\displaystyle\delta_{{a}}={\left|{\frac{{-{7.45}+{7.60787}}}{{-{7.60787}}}}\right|}{100}\%$$

$$={2.057}\%$$

For the value obtained in part (b), the percentage error is $$\displaystyle\delta_{{d}}={\left|{\frac{{-{7.478}+{7.60787}}}{{-{7.60787}}}}\right|}{100}\%$$

$$={1.707}\%$$

Therefore, the percentage error in part (a) is 2.075% and that is part(b) is 1.707%