# Given the following function: f(x)=1.01e^{4x}-4.62e^{3x}-3.11e^{2x}+12.2e^{x} a) Use three-digit rounding frithmetic, the assumption that e^{1.53}=4.62, and the fact that e^{nx}=(e^{x})^{n} to evaluate f(1.53) 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 f(1.53)=-7.60787

Question
Polynomial arithmetic
Given the following function:
$$\displaystyle{f{{\left({x}\right)}}}={1.01}{e}^{{{4}{x}}}-{4.62}{e}^{{{3}{x}}}-{3.11}{e}^{{{2}{x}}}+{12.2}{e}^{{{x}}}$$
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 $$\displaystyle{f{{\left({1.53}\right)}}}=-{7.60787}$$

2020-10-20
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 $$\displaystyle{1.707}\%.$$

### Relevant Questions

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 $$\displaystyle{f{{\left({1.53}\right)}}}=-{7.60787}$$
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B. Find the slope at P of the tangent to C.
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$$\begin{array}{c|c} x & f(x) & f'(x)\\ \hline 0.30 & 0.29552 & 0.95534\\ 0.32 & 0.31457 & 0.94924\\ 0.35 & 0.34290 & 0.93937\\ \end{array}$$

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$$Q(x) = (((3x - 5)x + 1)x 3)x + 5$$
Try to evaluate P(2) and Q(2) in your head, using the forms given. Which is easier? Now write the polynomial
R(x) =x^{5} - 2x^{4} + 3x^{3} - 2x^{2} + 3x + 4\) in “nested” form, like the polynomial Q. Use the nested form to find R(3) in your head.
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Use the theorems on derivatives to find the derivatives of the following function:
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