 # Given the following function: f(x)=1.01e^(4x)-4.62e^(3x)-3.11e^(2x)+12.2e^(x) Bevan Mcdonald 2020-10-19 Answered

Given the following function:
$f\left(x\right)=1.01{e}^{4x}-4.62{e}^{3x}-3.11{e}^{2x}+12.2{e}^{x}$
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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Step 1
a) $f\left(x\right)=1.01{e}^{4x}-4.62{e}^{3x}-3.11{e}^{2x}+12.2{e}^{x}-1.99$
$f\left(x\right)=1.01{\left({e}^{x}\right)}^{4}-4.62{\left({e}^{x}\right)}^{3}-3.11{\left({e}^{x}\right)}^{2}+12.2{e}^{x}-1.99$
$f\left(1.53\right)=1.01{\left({e}^{1.53}\right)}^{4}-4.62{\left({e}^{1.53}\right)}^{3}-3.11{\left({e}^{1.53}\right)}^{2}+12.2{e}^{1.53}-1.99$
$=1.01{\left(4.62\right)}^{4}-4.62{\left(4.62\right)}^{3}-3.11{\left(4.62\right)}^{2}+12.2\left(4.62\right)-1.99$
$=1.01\left(455.583\right)-4.62\left(98.611\right)-3.11\left(21.344\right)+56.364-1.99$
$=460.139-455.583-66.380+54.374$
$=-7.45$
Therefore, the value of $f\left(1.53\right)$ obtained by this method is $-7.45.$
Step 2
b) The given function can be factorized as 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 $-7.478$
Step 3
c) Percentage error $\delta$ is given by

Here ${V}_{A}$ is the actual value observed and ${V}_{E}$ is the expected value which is -7.60787 in this case.
For the value obtained in part (a), the percentage error is

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