Please check my answers! Thank you so much!

Step-by-step explanation:

1. f(x) = cos(x) → f(π/3) = ½

f'(x) = -sin(x) → f'(π/3) = -½√3

P₁(x) = f(a) + f'(a)/1! (x − a)

P₁(x) = ½ − ½√3 (x − π/3)

2. f(x) is a third order polynomial, so any third order Taylor series will equal f(x).

The remainder is the difference between f(x) and the partial sum.  So R₅ = 0 and R₂ = -5x² + 4x³.

I and II only

3. eˣ = 1 + x/1! + x²/2! + x³/3! + ...

e^(0.1) = 1 + 0.1/1! + 0.1²/2! + 0.1³/3! + ...

0.1ⁿ⁺¹/(n+1)! < 0.00001

By trial and error:

0.1¹⁺¹/(1+1)! = 0.005

0.1²⁺¹/(2+1)! = 0.00017

0.1³⁺¹/(3+1)! = 0.000004

n = 3

4. f(x) = ∑ aₙ xⁿ

g(x) = ∫₀ˣ f(t) dt

g(x) = ∫₀ˣ a₀ dt + ∫₀ˣ a₁t dt + ∫₀ˣ a₂t² dt + ... + C

g(x) = a₀x + a₁x²/2 + a₂x³/3 + ... + g(0)

g(x) = 3 + ∑ aₙ xⁿ⁺¹/(n+1)

5. f(x) = √x = x^½ → f(4) = 2

f'(x) = ½ x^(-½) → f'(4) = ¼

f''(x) = -¼ x^(-³/₂) → f''(4) = -¹/₃₂

f⁽³⁾(x) = ⅜ x^(-⁵/₂) → f⁽³⁾(4) = ³/₂₅₆

f⁽⁴⁾(x) = -¹⁵/₁₆ x^(-⁷/₂) → f⁽⁴⁾(4) = -¹⁵/₂₀₄₈

P₄(x) = f(4) + f'(4) (x−4) / 1! + f''(4) (x−4)² / 2! + f⁽³⁾(4) (x−4)³ / 3! + f⁽⁴⁾(4) (x−4)⁴ / 4!

P₄(x) = 2 + ¼ (x−4) − ¹/₆₄ (x−4)² + ¹/₅₁₂ (x−4)³ − ⁵/₁₆₃₈₄ (x−4)⁴

P₄(4.2) = 2 + ¼ (0.2) − ¹/₆₄ (0.2)² + ¹/₅₁₂ (0.2)³ − ⁵/₁₆₃₈₄ (0.2)⁴

P₄(4.2) = 2.049390137

a

step-by-step explanation:

when a number is between two straight lines it means the absolute value.

to find absolute value you need to know how many numbers is that number away from 0.

when finding absolute value the number will always come out positive

so absolute value of -3 is 3

absolute value of 3/2 is 3/2

absolute value of -2 is 2

absolute value of 3.5 is 3.5

absolute value of -1 is 1

now put these numbers in order:

a)   |-1|, |3/2|, |-2|, |-3|, |3.5|