Given information:

  f(x)=x,a=4,n=2,4≤x≤4.2

Formula Used:

Taylor Polynomials

  Tn(x)=∑i=0nf(i)(a)i!(x−a)i=f(a)+f'(a)1!(x−a)+f''(a)(x−a)22!…+fn(a)n!(x−a)n

Calculation:

Taylor polynomials Tn(x)

  f(x)=x,a=4,n=2,4≤x≤4.2T2(x)=f(a)+f'(a)1!(x−a)+f''(a)(x−a)22!f(x)=x,a=4f(4)=0f'(4)=14f''(4)=−132T2(x)=f(a)+f'(a)1!(x−a)+f''(a)(x−a)22!T2(x)=f(4)+f'(4)1!(x−4)+f''(4)(x−4)22!T2(x)=2+14(x−4)−(x−4)264

Taylor polynomials T2(x)

  T2(x)=2+14(x−4)−(x−4)264

Given information:

  f(x)=x,a=4,n=2,4≤x≤4.2

Formula Used:

  Tn(x)→f(x) as n→∞

And Tn(x) can be used as an approximation to

  f:f(x)=Tn(x)

Taylor’s Inequality, which says that if

  |f(n+1)(x)|≤M , then

  |Rn(x)|≤M(n+1)!|x−a|n+1

Calculation:

Taylor’s Inequality with a=4 and n = 2

  |R2(x)|≤M3!|x−a|3

Where |f3(x)|<M because x≤4.2

  |f3(x)|≤38(4)52|f3(x)|≤0.011718|R2(x)|≤0.0117183!|4.2−4|3|R2(x)|≤1.5625×10−5

The approximation is accurate to within

  1.5625×10−5

Given information:

  f(x)=x,a=4,n=2,4≤x≤4.2

  T2(x)=2+14(x−4)−(x−4)264

Formula Used:

  |Rn(x)|=|f(x)−Tn(x)|

Calculation:

  |R2(x)|=|f(x)−T2(x)|R2(x)=x−2−(x−4)4+(x−4)264