Let f be a function that has derivatives of all orders for all real numbers. In particular,assume f(1) = 3, f′(1) = −2, f′′(1) = 2, f′′′(1) = 4, and |f(4)(z)| = 2 for 0 < z < 0.1

Question

Let f be a function that has derivatives of all orders for all real numbers. In particular,assume f(1) = 3, f′(1) = −2, f′′(1) = 2, f′′′(1) = 4, and |f(4)(z)| = 2 for 0 < z < 0.1

(a) Write the third-degree Taylor Polynomial for f about x = 1 and use it to approximate f(1.1)

(b) Determine the accuracy of the previous approximation

(c) Write the second-degree Taylor Polynomial for f′ about x = 1 and use it to approximatef′(1.1)

Answer 1

(a) Third-Degree Taylor Polynomial for f(x) about x=1:

The formula for the third-degree Taylor polynomial is:

T₃(x) = f(1) + f'(1)(x-1) + f''(1)(x-1)² / 2! + f'''(1)(x-1)³ / 3!

Given values: f(1) = 3, f′(1) = −2, f′′(1) = 2, f′′′(1) = 4.

T₃(x) = 3 - 2(x-1) + (2/2)(x-1)² + (4/6)(x-1)³

T₃(x) = 3 - 2(x-1) + (x-1)² + (2/3)(x-1)³

Approximate: f(1.1), meaning x=1.1

T₃(1.1) = 3 - 2(1.1-1) + (1.1-1)² + (2/3)(1.1-1)³

T₃(1.1) = 3 - 2(0.1) + (0.1)² + (2/3)(0.1)³ = 3 - 0.2 + 0.01 +(2/3)(0.001) ≈ 2.8106667

(b) Accuracy of the Approximation is bounded by the Langrange Remainder:

R₃(x) = (|f⁽ⁿ⁺¹⁾(z)| / (n+1)! )(x-a)⁽ⁿ⁺¹⁾

R₃(x) =( |f⁽⁴⁾(z)| / 4! ) • | x - 1|⁴

and |f⁽⁴⁾(z)| ≤ 2.

R₃(1.1) ≤ ( 2 / 4! ) • | 1.1 - 1|⁴ = (2 / 24)(0.1)⁴ =(1/12) (.0001)

R₃(1.1) ≤ 0.00000833.

(c) Second-Degree Taylor Polynomial for f′(x) about x=1.

T'₂(x) = f'(x)+f''(1)(x-1)+f'''(1)/2! (x-1)².

T'₂(1.1) = f'(1.1)+f''(1)(1.1-1)+f'''(1)/2! (1.1-1)²

T'₂(1.1) = -2 + 2(0.1)+4/2 (0.1)² = -2 + 2(0.1)+ 2 (0.1)²

T'₂(1.1) = -2 + 0.2 + 0.02 = -1.78

So, the approximation is: f'(1.1) ≈ -1.78