a) The #nth# degree Taylor Polynomial of a function #f(x)# about #x=a# is defined as
#T_n(x)=f(a)+f'(a)(x-a)+f''(a)(x-a)^2/(2!)
+f^((3))(a)(x-a)^3/(3!)+...+f^((n))(a)(x-a)^n/(n!)#
So, to find #T_3(x)# for #f(x)=x^-3# at #a=1,# we need the #f(1)# and the first three derivatives each evaluated at #1#.
#f(1)=1^-3=1#
#f'(x)=-3x^-4, f'(1)=-3#
#f''(x)=12x^-5, f''(1)=12#
#f^((3))(x)=-60x^-6, f^((3))(1)=-60#
Thus,
#T_3(x)=1-3(x-1)+12(x-1)^2/(2!)-60(x-1)^3/(3!)#
#T_3(x)=1-3(x-1)+12(x-1)^2/(2)-60(x-1)^3/(6)#
#T_3(x)=1-3(x-1)+6(x-1)^2-10(x-1)^3#
b) We can find #R_3(x)# using Taylor's Inequality, which tells us that
#|R_n(x)|<=C|x-a|^(n+1)/((n+1)!)#
Where #C# is the bounding value for the #(n+1)th# derivative on the interval #2.5<=x<=3.5# .
That is, #C>=|f^((n+1))(x)|# on #2.5<=x<=3.5#
We're looking to maximize #f^((n+1))(x)# on #2.5<=x<=3.5#.
Well, we have #n=3, n+1=4, f^((4))(x)=360x^-7#
#f^((4))(x)=360/x^7#
On #2.5<=x<=3.5, |360/x^7|=360/x^7# is maximized for #x=2.5,# which yields the smallest possible denominator on this interval and therefore the largest value overall.
So, #C=|f^((4))(2.5)|=360/(2.5)^7=0.589824#
Thus,
#|R_3(x)|<=C|x-1|^(3+1)/((3+1)!)#
#|R_3(x)|<=0.589824|x-1|^(4)/(4!)#
We know
#2.5<=x<=3.5#
So,
#2.5-1<=x-1<=3.5-1#
#1.5<=x-1<=2.5#
We see #|x-1|<=2.5#, so #|x-1|^4<=(2.5)^4=39.0625#
Then,
#|R_3(x)|<=0.589824*39.0625/(4!)=0.9599#
#R_3(x)<=0.9599#
