Given the following function: f(x)=1.01e^(4x)-4.62e^(3x)-3.11e^(2x)+12.2e^(x)

Step 1
a) ${\displaystyle f\left(x\right)=1.01e^{4x}-4.62e^{3x}-3.11e^{2x}+12.2e^x-1.99}$
${\displaystyle f\left(x\right)=1.01{\left(e^x\right)}^4-4.62{\left(e^x\right)}^3-3.11{\left(e^x\right)}^2+12.2e^x-1.99}$
${\displaystyle f\left(1.53\right)=1.01{\left(e^{1.53}\right)}^4-4.62{\left(e^{1.53}\right)}^3-3.11{\left(e^{1.53}\right)}^2+12.2e^{1.53}-1.99}$
${\displaystyle =1.01{\left(4.62\right)}^4-4.62{\left(4.62\right)}^3-3.11{\left(4.62\right)}^2+12.2\left(4.62\right)-1.99}$
${\displaystyle =1.01\left(455.583\right)-4.62\left(98.611\right)-3.11\left(21.344\right)+56.364-1.99}$
${\displaystyle =460.139-455.583-66.380+54.374}$
${\displaystyle =-7.45}$
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.
${\displaystyle f\left(x\right)=1.01(e^x+1.715)(e^x-0.173)(e^x-1.415)(e^x-4.702)}$
On substituting ${\displaystyle x=1.53}$ and using ${\displaystyle e^{1.53}=4.62}$, we get
${\displaystyle f\left(1.53\right)=1.01(4.62+1.715)(4.62-0.173)(4.62-1.415)(4.62-4.702)}$
${\displaystyle =1.01\left(6.335\right)\left(4.447\right)\left(3.205\right)(-0.082)}$
${\displaystyle =-7.478}$
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
${\displaystyle {\delta }_a=\left|\frac{-7.45+7.60787}{-7.60787}\right|100\mathrm{\%}}$

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