Consider f x 5 lnx xSolutionAnswers Series for fx f x 5 l

Consider f (x) = 5 ln(x) ? x.

Answers.

Series for f(x);

f (x) = 5 ln(x) - x

f (1) = 5 ln 1 - 1 = -1

f ´(x) = 5/x - 1

f ´(1) = 5 - 1 = 4

f ´´(x) = -5/x²

f ´´(1) = -5/1 = -5

so,

T₂ (x) = f(1) + f ´(1) (x-1) + f ´´(1)/2! (x-1)²

b) Using QAEB for |f(x) - T₂| ;

| [5 ln(x) - x] - [-1 + 4(x-1)-5/2!(x-1)²] | <= 10/6(1-a)³ = 5/3(1-a)³

c) T₂ (x) <= 0.36

replacing;

5/3(1-a)³ <= 0.36

5/3x0.36 = (1 - a)³

cubic root for both sides, we have;

a - 1 = 0.8434

a = 1.8434