Respuesta :
Compute the necessary values/derivatives of [tex]f(x)[/tex] at [tex]x=1[/tex]:
[tex]f(1)=3[/tex]
[tex]f'(1)=2[/tex]
[tex]f''(1)=-12[/tex]
[tex]f'''(1)=-12[/tex]
[tex]f^{(4)}(1)=48[/tex]
[tex]f^{(5)}(1)=120[/tex]
Taylor's theorem then says we can "approximate" (in quotes because the Taylor polynomial for a polynomial is another, exact polynomial) [tex]f(x)[/tex] at [tex]x=1[/tex] by
[tex]T_5(x)=\dfrac3{0!}+\dfrac2{1!}(x-1)-\dfrac{12}{2!}(x-1)^2-\dfrac{12}{3!}(x-1)^3+\dfrac{48}{4!}(x-1)^4+\dfrac{120}{5!}(x-1)^5[/tex]
[tex]T_5(x)=3+2(x-1)-6(x-1)^2-2(x-1)^3+2(x-1)^4+(x-1)^5[/tex]
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Another way of doing this would be to solve for the coefficients [tex]a,b,c,d,e,g[/tex] in
[tex]f(x)=a+b(x-1)+c(x-1)^2+d(x-1)^3+e(x-1)^4+g(x-1)^5[/tex]
by expanding the right hand side and matching up terms with the same power of [tex]x[/tex].
The coefficients are 3, 2, -6, -2, 2, and 1 and this can be determined by using the series formula.
Given :
- [tex]\rm f(x) = x^5-3x^4+2x^2+5x-2[/tex]
- b= 1
- [tex]\rm T_5(x) = (x -1)^1+ (x-1)^2+ (x-1)^3+ (x-1)^4+ (x-1)^5[/tex]
The formula of the series is given below:
[tex]\rm T(x) = \sum^{\infty}_{n=0} \dfrac{f^n(b)}{n!}(x-b)^n[/tex]
Now, determine the value of [tex]\rm f'(1)[/tex].
[tex]\rm f'(x) = 5x^4-12x^3+4x+5[/tex]
[tex]\rm f'(1) = 5 - 12 + 4 + 5 = 2[/tex]
Now, determine the value of [tex]\rm f''(1)[/tex].
[tex]\rm f''(x) = 20x^3-36x^2+4[/tex]
[tex]\rm f"(1) = 20 - 36 + 4 = -12[/tex]
Now, determine the value of [tex]\rm f'''(1)[/tex].
[tex]\rm f'''(x) = 60x^2-72x[/tex]
[tex]\rm f'''(1) = 60 - 72 = - 12[/tex]
Now, determine the value of [tex]\rm f''''(1)[/tex].
[tex]\rm f''''(x) = 120x-72[/tex]
[tex]\rm f''''(1) = 48[/tex]
The coefficients are 3, 2, -6, -2, 2, and 1 and this can be determined by using the series formula.
For more information, refer to the link given below:
https://brainly.com/question/795909
