Step 1

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

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

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

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

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

\(\displaystyle={460.139}-{455.583}-{66.380}+{54.374}\)

\(\displaystyle=-{7.45}\)

Therefore, the value of \(\displaystyle{f{{\left({1.53}\right)}}}\) obtained by this method is \(\displaystyle-{7.45}.\)

Step 2

b) The given function can be factorized as 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

\(\displaystyle{f{{\left({1.53}\right)}}}={1.01}{\left({4.62}+{1.715}\right)}{\left({4.62}-{0.173}\right)}{\left({4.62}-{1.415}\right)}{\left({4.62}-{4.702}\right)}\)

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

\(\displaystyle=-{7.478}\)

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

Step 3

c) Percentage error \(\displaystyle\delta\) is given by

\(\displaystyle\delta={\left|{\frac{{{V}_{{{A}}}\ -\ {V}_{{{E}}}}}{{{V}_{{{E}}}}}}\right|}\)

Here \(\displaystyle{V}_{{{A}}}\) is the actual value observed and \(\displaystyle{V}_{{{E}}}\) is the expected value which is -7.60787 in this case.

For the value obtained in part (a), the percentage error is

\(\displaystyle\delta_{{{a}}}={\left|{\frac{{-{7.45}+{7.60787}}}{{-{7.60787}}}}\right|}\ {100}\%\)

\(\displaystyle={2.075}\%\)

For the value obtained in part (b), the percentage error is

\(\displaystyle\delta_{{{b}}}={\left|{\frac{{-{7.478}+{7.60787}}}{{-{7.60787}}}}\right|}\ {100}\%\)

\(\displaystyle={1.707}\%\)

Therefore, the percentage error in part (a) is \(\displaystyle{2.075}\%\) and that in part (b) is \(\displaystyle{1.707}\%.\)

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

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

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

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

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

\(\displaystyle={460.139}-{455.583}-{66.380}+{54.374}\)

\(\displaystyle=-{7.45}\)

Therefore, the value of \(\displaystyle{f{{\left({1.53}\right)}}}\) obtained by this method is \(\displaystyle-{7.45}.\)

Step 2

b) The given function can be factorized as 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

\(\displaystyle{f{{\left({1.53}\right)}}}={1.01}{\left({4.62}+{1.715}\right)}{\left({4.62}-{0.173}\right)}{\left({4.62}-{1.415}\right)}{\left({4.62}-{4.702}\right)}\)

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

\(\displaystyle=-{7.478}\)

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

Step 3

c) Percentage error \(\displaystyle\delta\) is given by

\(\displaystyle\delta={\left|{\frac{{{V}_{{{A}}}\ -\ {V}_{{{E}}}}}{{{V}_{{{E}}}}}}\right|}\)

Here \(\displaystyle{V}_{{{A}}}\) is the actual value observed and \(\displaystyle{V}_{{{E}}}\) is the expected value which is -7.60787 in this case.

For the value obtained in part (a), the percentage error is

\(\displaystyle\delta_{{{a}}}={\left|{\frac{{-{7.45}+{7.60787}}}{{-{7.60787}}}}\right|}\ {100}\%\)

\(\displaystyle={2.075}\%\)

For the value obtained in part (b), the percentage error is

\(\displaystyle\delta_{{{b}}}={\left|{\frac{{-{7.478}+{7.60787}}}{{-{7.60787}}}}\right|}\ {100}\%\)

\(\displaystyle={1.707}\%\)

Therefore, the percentage error in part (a) is \(\displaystyle{2.075}\%\) and that in part (b) is \(\displaystyle{1.707}\%.\)