 # 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 Suman Cole 2020-11-01 Answered
Given the following function: $f\left(x\right)=1.01{e}^{4x}-4.62{e}^{3x}-3.11{e}^{2x}+12.2{e}^{x}-1.99$ a)Use three-digit rounding frithmetic, the assumption that ${e}^{1.53}=4.62$, and the fact that ${e}^{nx}={\left({e}^{x}\right)}^{n}$ to evaluate $f\left(1.53\right)$ 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 $f\left(1.53\right)=-7.60787$
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a) $f\left(x\right)=1.01{e}^{4x}-4.62{e}^{3}x-3.11{e}^{2x}+12.2{e}^{x}-1.99$

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

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

$=460.139-455.583-66.380+54.374$

$=-7.45$

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

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

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

$=-7.478$

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

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

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