Taylor's Series

[Guest]

Hello can someone help me out these two problems. lol k. : ]. I am pretty stuck, and I don't know where to go with these two problems and how to solve it at all. = (.

For the second one, I already know how to do a for finding 0.3, however I am stuck with b. How do you do that? lol k. : ]. And also the rest. Sorry but thanks in advance. lol k. : ]. I apperciate your help. lol k. : ]. Thanks. lol k. : ].

Thanks in advance. I sincerely apperciate it. lol k. : ].

 

[soroban]

Hello, atse1900!

Here is most of the first one . . .

The Taylor series about $\displaystyle x\,=\,0$ for a a certain function $\displaystyle f$ converges for all $\displaystyle x$ in the interval of convergence.

\(\displaystyle \text{The }n^{th}\text{ derivative of }f\text{ at }x\,=\,0\text{ is: }\; f^{(n)}(0)\;=\;\frac{(-1)^{n+1}(n\,+\,1)!}{5^n(n\,-\,1)^2\;\) for \(\displaystyle n\,\geq\,2\).

The graph of $\displaystyle f$ has a horizontal tangent line at $\displaystyle x\,=\,0$ and $\displaystyle f(0)\,=\,6$.

a) Determine whether $\displaystyle f$ has a relative maximum, minimum, or neither at $\displaystyle x\,=\,0$.
$\displaystyle \;\;\;$Justify your answer.

b) Write the third-degree Taylor polynomial for $\displaystyle f$ about $\displaystyle x\,=\,0$.

Use the Second Derivative Test.

For $\displaystyle n\,=\,2:\;\;f''(0)\;=\;\frac{(-1)^3(3!)}{5^2(1^2)}\;=\;-\frac{6}{25}$

The second derivative is negative at $\displaystyle x\,=\,0$; the graph is concave down: $\displaystyle \,\cap$

Therefore, there is a relative maximum at $\displaystyle (0,\,6)$.


For part (b), let $\displaystyle n\,=\,0,\,1,\,2,\,3$ and construct the Taylor Series.