Question

f(x) = e^(2x) + e^(−2x)
Find an upper bound for the error in using the second degree Maclaurin polynomial of f to approximate f(0.5).

Answer 1

I guess I have to use the Lagrange remainder formula, right? \[E _{n}(x) = f(x) - T _{n}(x) =\frac{ f ^{(n+1)}(c) }{ (n+1)! }(x-a)^{(n+1)}\]
Where c lies between x and a

Answer 2

yes
just plug the rest in

Answer 3

So here a=0 right?

Answer 4

i think so

Answer 5

Is my 3rd derivative correct? I got... \[8e ^{2x}-8e ^{-2x}\]

Answer 6

ya

Answer 7

Cool, so then I get \[E _{2}(0.5)=\frac{ e ^{2c} -e ^{-2c}}{ 6 }\]

Answer 8

yes! im surprised u asked for help cuz u get it

Answer 9

Lol I don't know what to do next :D

Answer 10

that is f(x)

Answer 11

i see what u do now

Answer 12

u have to

Answer 13

replace x with .5

Answer 14

What if I was trying to find the lower bound?