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

Solution :

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

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

    $\frac{f""(2)}{2!} =-9$  , 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 $=\frac{6}{4!}|0-2|^4= 4$

     $f(0) \leq T(0)+4 = -1$

     Therefore, $f(0)$ is negative.