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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.

Sagot :

Solution :

Given : [tex]$T(x)=7-9(x-2)^2-3(x-2)^3$[/tex]

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

    [tex]$\frac{f""(2)}{2!} =-9$[/tex]  , so f''(2) = [tex]$-18$[/tex]

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

   Since f''(2) = [tex]$-18$[/tex] < 0, [tex]$f(2)$[/tex] is relative maximum value.

c).  [tex]$f(0)=T(0)=-5$[/tex]

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

d). The Lagrange [tex]$\text{error}$[/tex] bound [tex]$=\frac{6}{4!}|0-2|^4= 4$[/tex]

     [tex]$f(0) \leq T(0)+4 = -1$[/tex]

     Therefore, [tex]$f(0)$[/tex] is negative.

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