Question

Let f be a function having derivatives of all orders for allreal numbers. The third-degree Taylor polynomial for f about x=2 isgiven by

T(x) = 7-9(x-2)2-3(x-2)3


a. find f(2) and f '' (2).


b. Is there enough information given to determine wheather fhas a critical point at x = 2?


If not, explain why not.


If so, determine whether f(2) is a relative maximum, arelative minimum, or neither, and justify your answer.


c. Use T(x) to find an approximation for f(0). Is there enoughinformation given to determine whether f has a critical point atx=0?


If not, explain why not.


If so, determine whether f (0) is a relative maximum, arelative minimum, or neither, and justify your answer.


d. The fourth derivative of f satisfies the inequality for all x in the closed interval [0,2]. Use thelagrange error bound on the approximation to f (0) found in part(c) to explain why f (0) is negative.

Answer 1

Given :
`$T(x)=7-9(x-2)^2-3(x-2)^3$`

a). f(2) = T(2) = 7

`$(f` , so f''(2) =
`$-18$`

b). Yes, since f'(2) = T'(2)
`$=$` 0,
`$f$` does have the critical point at
`$x=2.$`

Since f''(2) =
`$-18$` < 0,
`$f(2)$` is relative maximum value.

c).
`$f(0)=T(0)=-5$`

It is also not possible for determining if
`$f$` has a critical point at x = 0 because
`$T(x)$` gives exact information only at
`$x=2.$`

d). The Lagrange
`$\text{error}$` bound
`$=(6)/(4!)|0-2|^4= 4$`

`$f(0) \leq T(0)+4 = -1$`

Therefore,
`$f(0)$` is negative.