`f(x)=4/(3x+2) , c=3` Find a power series for the function, centered at c and determine the interval of convergence.

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To determine the power series centered at c, we may apply the formula for Taylor series:

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

To list the `f^n(x)` for the given function `f(x)=4/(3x+2)` centered at `c=2` , we may apply Law of Exponent: `1/x^n =...

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To determine the power series centered at c, we may apply the formula for Taylor series:

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

To list the `f^n(x)` for the given function `f(x)=4/(3x+2)` centered at `c=2` , we may apply Law of Exponent: `1/x^n = x^-n`  and  Power rule for derivative: `d/(dx) x^n= n *x^(n-1)` .

`f(x) =4/(3x+2)`

     ` =4(3x+2)^(-1)`

Let `u =3x+2` then `(du)/(dx) = 3` .

`d/(dx) c*(3x+2)^n = c *d/(dx) (3x+2)^n`

                            `= c *(n* (3x+2)^(n-1)*3`

                            ` = 3cn(3x+2)^(n-1)`

`f'(x) =d/(dx) 4(3x+2)^(-1)`

          `=3*4*(-1) *(3x+2)^(-1-1)`

           `=-12(3x+2)^(-2) or 2/(3x+2)^2`

`f^2(x) =d/(dx) -12(3x+2)^(-2)`

           `=3*(-12)(-2)(3x+2)^(-2-1)`

          `=72(3x+2)^(-3) or 72/(3x+2)^3`

`f^3(x) =d/(dx) 72(3x+2)^(-3)`

           `=3*(72)(-3)(3x+2)^(-3-1)`

           `=-648(3x+2)^(-4) or -648/(3x+2)^4`

Plug-in `x=3` for each `f^n(x)` , we get:

`f(3)=4/(3(3)+2)`

        `=4/ 11`

`f'(3)=-12/(3(3)+2)^2`

          `=-12/11^2`

          `= -12/121`

`f^2(3)=72/(3(3)+2)^3 `

           `=72/11^3`

           `=72/1331`

`f^3(3)=-648/(3(3)+2)^4 `

           `=-648/11^4`

           `= -648/14641`

Plug-in the values on the formula for Taylor series, we get:

`4/(3x+2)= sum_(n=0)^oo (f^n(3))/(n!) (x-3)^n`

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

` =4/11+(-12/121)(x-3) +(72/1331)/(2!)(x-3)^2 +(-648/14641)/(3!)(x-3)^3 +...`

` =4/11-(12/121)(x-3) +(72/1331)/2(x-3)^2 - (648/14641)/6(x-3)^3 +...`

` =4/11-12/121(x-3) +36/1331(x-3)^2 -108/14641(x-3)^3 +...`

` = sum_(n=1)^oo 4(-3(x-3))^(n-1)/11^n`

` = sum_(n=1)^oo 4(-3(x-3))^(-1)(-3(x-3))^n/11^n`

` = sum_(n=1)^oo 4/(-3(x-3))((-3(x-3))/11)^n`

` =sum_(n=1)^oo 4/(-3x+9)((-3(x-3))/11)^n`

To determine the interval of convergence, we may apply geometric series test wherein the series `sum_(n=0)^oo a*r^n`  is convergent if `|r|lt1 or -1 ltrlt 1` . If `|r|gt=1` then the geometric series diverges.

By comparing  `sum_(n=1)^oo 4/(-3x+9)((-3(x-3))/11)^n` with `sum_(n=0)^oo a*r^n` , we determine: `r = (-3(x-3))/11` .

Apply the condition for convergence of geometric series: `|r|lt1` .

`|(-3(x-3))/11|lt1`

`|-1|*|(3(x-3))/11|lt1`

`1*|(3(x-3))/11|lt1`

`|(3(x-3))/11|lt1`

`|(3x-9)/11|lt1`

`-1lt(3x-9)/11lt1`

Multiply each sides by `11` :

`-1*11lt(3x-9)/11*11lt1*11`

`-11lt3x-9lt11`

Add 9 on each sides:

`-11+9lt3x-9+9lt11+9`

`-2lt3xlt20`

Divide each side by `3` :

`-2/3lt(3x)/3lt20/3`

 

`-2/3ltxlt20/3 `

Thus, the power series  of the function `f(x) =4/(3x+2)` centered at `c=3` is `sum_(n=1)^oo 4/(-3x+9)((-3(x-3))/11)^n`

 with an interval of convergence: `-2/3ltxlt20/3` .

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