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.