a)
\(\displaystyle{f{{\left({x}\right)}}}={1.01}{e}^{{{4}{x}}}-{4.62}{e}^{{3}}{x}-{3.11}{e}^{{{2}{x}}}+{12.2}{e}^{{x}}-{1.99}{N}{S}{K}{f{{\left({x}\right)}}}={1.01}{\left({e}^{{x}}\right)}^{{4}}-{4.62}{\left({e}^{{x}}\right)}^{{3}}-{3.11}{\left({e}^{{{x}}}\right)}^{{2}}-{12.2}{e}^{{x}}-{1.99}{\left({a}{s}{e}^{{{n}{x}}}={\left({e}^{{x}}\right)}^{{n}}\right)}{N}{S}{K}{f{{\left({1.53}\right)}}}={1.01}{\left({e}^{{{1.53}}}\right)}^{{4}}-{4.62}{\left({e}^{{{1.53}}}\right)}^{{3}}-{3.11}{\left({e}^{{{1.53}}}\right)}^{{2}}+{12.2}{e}^{{{1.53}}}-{1.99}{N}{S}{K}={1.01}{\left({4.62}\right)}^{{4}}-{4.62}{\left({4.62}\right)}^{{3}}-{3.11}{\left({4.62}\right)}^{{2}}+{12.2}{\left({4.62}\right)}-{1.99}{\left({a}{s}{e}^{{{1.53}={4.62}}}\right)}{N}{S}{K}={1.01}{\left({455.583}\right)}-{4.62}{\left({98.611}\right)}-{3.11}{\left({21.344}\right)}+{56.364}-{1.99}{N}{S}{K}={460.139}-{455.583}-{66.380}+{54.374}{N}{S}{K}=-{7.45}\)
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
\(\displaystyle{f{{\left({1.53}\right)}}}={1.01}{\left({4.62}+{1.715}\right)}{\left({4.62}-{0.173}\right)}{\left({4.62}-{1.415}\right)}{\left({4.62}-{4.702}\right)}{N}{S}{K}={1.01}{\left({6.335}\right)}{\left({4.447}\right)}{\left({3.205}\right)}{\left(-{0.082}\right)}{N}{S}{K}=-{7.478}\)
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
\(\displaystyle\delta_{{a}}={\left|{\frac{{-{7.45}+{7.60787}}}{{-{7.60787}}}}\right|}{100}\%{N}{S}{K}={2.057}\%\)
For the value obtained in part (b), the percentage error is
\(\displaystyle\delta_{{d}}={\left|{\frac{{-{7.478}+{7.60787}}}{{-{7.60787}}}}\right|}{100}\%{N}{S}{K}={1.707}\%\)
Therefore, the percentage error in part (a) is 2.075% and that is part(b) is 1.707%