Question

Taylor series expansion

Answer 1

Can I get help with 3?

Answer 2

Oh, and the new function f(x) involving e is the one you use for #3, DON'T USE THE FIRST FUNCTION THEY GAVE

Answer 3

@ganeshie8

Answer 4

Do I plug in (x^2+2) into e^-2x^2?

Answer 5

Are we supposed to find the derivatives freshly for f(x) = e^(-2x^2)
or use the values given at the start of question ?

Answer 6

I find the question a bit ambiguous..

Answer 7

Either ways, I think you just need to show that g(x) meets below for having relative minimum at x=0 :

1) g'(0)=0
2) g''(0)>0

Answer 8

Yes, but I stiill have to show it

Answer 9

Wouldn't it be \[\Large g(x)=f(x^2+2)=e^{{-2(x^2+2)}^{2}}\]

Answer 10

I just think it will get ugly if you try to write a taylor series expansion of that

Answer 11

yeah
http://www.wolframalpha.com/input/?i=maclaurin+series+e%5E%28-2%28x%5E2%2B2%29%5E2%29

Answer 12

But you can see that there is no x term, there fore g'(0)=0 and then the coeffcient of the x^2 term is -8/e^8

f"(0) then equals -2!8/e^-8 but tht is negative so it concave down which is a maximum

Answer 13

Yes, x=0 gives a relative maximum. question must be wrong

Answer 14

ARGH.

Also, They tell me to show work. And um... this is the solution. My teacher or whoever writes this stuff must not check the questions


http://www.emathhelp.net/calculators/calculus-1/taylor-and-maclaurin-series-calculator/?f=e%5E%28-2%28x%5E2%2B2%29%5E2%29&p=0&n=4&steps=on

Answer 15

Yeah mistakes in questions are frustrating

Answer 16

You have seen a couple of my problems. Bad communication etc. This frustrates me but oh well. This is this the last lesson really.