approximate $f(3.63)$ when $f(x)=x^{3}e^{x - 1}$ and $h = 0.25$ using backward fdd. round your answer up to…

Explanation:

Step1: Recall the backward - finite - difference formula for the second - derivative

The backward - finite - difference formula for the second - derivative is $f''(x)\approx\frac{f(x)-2f(x - h)+f(x - 2h)}{h^{2}}$.

Step2: Calculate $x - h$ and $x - 2h$

Given $x = 3.63$ and $h=0.25$. Then $x - h=3.63 - 0.25 = 3.38$ and $x - 2h=3.63-2\times0.25=3.63 - 0.5 = 3.13$.

Step3: Calculate $f(x)$, $f(x - h)$ and $f(x - 2h)$

If $f(x)=x^{3}e^{x - 1}$, then:

• $f(3.63)=(3.63)^{3}e^{3.63 - 1}=(3.63)^{3}e^{2.63}$.
  • $(3.63)^{3}=3.63\times3.63\times3.63 = 47.83$.
  • $e^{2.63}\approx13.89$. So $f(3.63)=47.83\times13.89 = 664.36$.
• $f(3.38)=(3.38)^{3}e^{3.38 - 1}=(3.38)^{3}e^{2.38}$.
  • $(3.38)^{3}=3.38\times3.38\times3.38 = 38.61$.
  • $e^{2.38}\approx10.75$. So $f(3.38)=38.61\times10.75 = 415.06$.
• $f(3.13)=(3.13)^{3}e^{3.13 - 1}=(3.13)^{3}e^{2.13}$.
  • $(3.13)^{3}=3.13\times3.13\times3.13 = 30.66$.
  • $e^{2.13}\approx8.42$. So $f(3.13)=30.66\times8.42 = 258.16$.

Step4: Apply the backward - finite - difference formula

$f''(3.63)\approx\frac{f(3.63)-2f(3.38)+f(3.13)}{(0.25)^{2}}$. Substitute the values: [ \begin{align*} &\frac{664.36-2\times415.06 + 258.16}{0.0625}\ =&\frac{664.36-830.12+258.16}{0.0625}\ =&\frac{(664.36 + 258.16)-830.12}{0.0625}\ =&\frac{922.52-830.12}{0.0625}\ =&\frac{92.4}{0.0625}\ =&1478.40 \end{align*} ]

Answer:

$1478.40$