`f(x)=root(3)(x),n=3,c=8` Find the n'th Taylor Polynomial centered at c

Taylor series is an example of infinite series derived from the expansion of `f(x)` about a single point. It is represented by infinite sum of `f^n(x)`  centered at `x=c ` .The general formula for Taylor series is:

`f(x) = sum_(n=0)^oo (f^n(c))/(n!) (x-c)^n`

or

`f(x) = f(c) + f'(c) (x-c)+ (f'(c))/(2!) (x-c)^2+ (f'(c))/(3!) (x-c)^3+ (f'(c))/(4!) (x-c)^4+...`

To evaluate the given function `f(x) =root(3)(x)` , we may express it in terms of fractional exponent using the radical property: `root(n)(x)= x^(1/n)` . The function becomes:

`f(x) = (x)^(1/3)` .

Apply the definition of the Taylor series by listing the `f^n(x)` up to `n=3` .

 We determine each derivative using Power Rule for differentiation: `d/(dx) x^n = n*x^(n-1)` .

`f(x) = (x)^(1/3)`

`f'(x) = 1/3 * x^(1/3-1)`

          `= 1/3x^(-2/3) or1/(3x^(2/3) )`

`f^2(x) = d/(dx) (1/3x^(-2/3))`

          `= 1/3 * d/(dx) (x^(-2/3))`

          `= 1/3*(-2/3x^(-2/3-1))`

          `= -2/9 x^(-5/3) or -2/(9x^(5/3))`

`f^3(x) = d/(dx) (-2/9 x^(-5/3))`

          `= -2/9*d/(dx) (x^(-5/3))`

         `= -2/9*(-5/3x^(-5/3-1))`

          `=10/27 x^(-8/3) or 10/(27x^(8/3))`

Plug-in x=8, we get:

`f(8) = (8)^(1/3)`

        `= 2`

`f'(8)=1/(3*8^(2/3) )`

        `=1/(3*4)`

        `=1/12`

`f^2(8)=-2/(9*8^(5/3))`

          `=-2/(9*32)`

          `= -2/288`

          `=-1/144`

`f^3(8)=10/(27*8^(8/3))`

          `=10/(27*256)`

          `= 10/6912`

          `=5/3456`

Applying the formula for Taylor series centered at `c=8` , we get:

`sum_(n=0)^3 (f^n(8))/(n!)(x-8)^n`

    `=f(8) + f'(8) (x-8)+ (f'(8))/(2!) (x-8)^2+ (f'(8))/(3!) (x-8)^3`

    `=2+ (1/12) (x-8)+ (-1/144)/(2!) (x-8)^2+ (5/3456)/(3!) (x-8)^3`

    `=2+ 1/12 (x-8)-1/(144*2) (x-8)^2+ 5/(3456*6) (x-8)^3 `

    `=2+ 1/12 (x-8)-1/288 (x-8)^2+ 5/20736 (x-8)^3 `

 The Taylor polynomial of degree `n=3 `  for the given function `f(x)=root(3)(x)`  centered at `c=8`  will be:

`P(x)=2+ 1/12 (x-8)-1/288 (x-8)^2+ 5/20736 (x-8)^3 `