Please send the answers within Half hour. Thanks. Follow the Steps:

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Please send the answers within Half hour. Thanks.

Please Send The Answers Within Half Hour Thanks Follow The Steps 1 (19.55 KiB) Viewed 20 times

Follow the Steps:

Please Send The Answers Within Half Hour Thanks Follow The Steps 2 (61.94 KiB) Viewed 20 times

Q3 (7 points) (20-²(x-2)x) Let f(x) = cos(x) + 8(x - 1) sin(x) 276 given that: 3(x-2) 3(x-2)² (x-2)³ (1) P₂ (T) = 10 76 is the cubic 274 672 Taylor polynomial of f(x) at x = 2. (2) The Taylor series of f(x) at x = 2 is NOT alternating. (3) ƒ(¹) (x) = (2x − x²) cos(TT) Find the maximum possible error in using P3(z) to approximate f(x) in the interval 0 ≤ ≤ 2. You are
2. Find the Taylor series about x = 1 for the function x-1 f(x) = (3x + 1)1/3* Write your final answer in sigma notation, simplified as much as possible. You must use a method that guarantees that the series converges to f(x). What is the radius of convergence of the series? 1 1 1 3(x-1) 4¹/3 4 (3x + 1)¹/3 [3(x − 1) + 4]¹/3 = - 5 [1+302-11-1 = { 1 + (-1/3) (-1)] + (-1/3)(-4/3)(x-1)] 41/3 2! 3 3 + (-1/3)(-4/3)(-7/3) 3! | (a = [1 − ( x − 1) + (1-4) ( (x-1)²- (1.4.7) 3! 4³ (x - 1)³- 4] 2! 4² ∞ (-1)"[1 4 7 (3n - 2)] 1+ ¹ (2-1)"} 4¹/3 n! 4n n=1 =47/3 + [ (-1) [1 -4.7 (3n-2)] n=1 (x-1) + 1) + Ĉ! 4¹/3 n=1 ∞ = -47/3 ( (x-1)+) 3 -2) < 1 = = 4¹/3 Thus, x-1 (3x + 1)¹/3 Since the series is valid for - (x - 1)" n! 4n+1/3 ... (-1)"[1.4.7 (3n-2)] n! 4n+1/3 -(x − 1) n+1 (-1)+¹[1-4-7 (3n — 5)], ... -(x - 1)". (n-1)! 4n-2/3 x2 < 4/3, the radius of converence is 4/3. n=2