f(x)=3/(2x-1) ,c=2 Find a power series for the function, centered at c and determine the interval of convergence.

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)=3/(2x-1) 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) =3/(2x-1)
           =3(2x-1)^(-1)
Let u =2x-1 then (du)/(dx) = 2
d/(dx) c*(2x-1)^n = c *d/(dx) (2x-1)^n
                           = c *(n* (2x-1)^(n-1)*2
                            = 2cn(2x-1)^(n-1)
f'(x) =d/(dx) 3(2x-1)^(-1)
            =2*3*(-1)(2x-1)^(-1-1)
            =-6(2x-1)^(-2) or 2/(2x-1)^2
f^2(x) =d/(dx) -6(2x-1)^(-2)
            =2*(-6)(-2)(2x-1)^(-2-1)
             =24(2x-1)^(-3) or 24/(2x-1)^3
f^3(x) =d/(dx) 24(2x-1)^(-3)
           =2*(24)(-3)(2x-1)^(-3-1)
           =-144(2x-1)^(-4) or -144/(2x-1)^4
Plug-in x=2 for each f^n(x), we get:
f(2)=3/(2(2)-1)
          =3/ 3
          =1
f'(2)=-6/(2(2)-1)^2
          =-6/3^2
          = -2/3
f^2(2)=24/(2(2)-1)^3
            =24/3^3
            =8/9
f^3(2)=-144/(2(2)-1)^4
            =-144/3^4
            = -16/9
Plug-in the values on the formula for Taylor series, we get:
3/(2x-1) = sum_(n=0)^oo (f^n(2))/(n!) (x-2)^n
= sum_(n=0)^oo (f^n(2))/(n!) (x-2)^n
=1+(-2/3)(x-2) +(8/9)/(2!)(x-2)^2 +(-16/9)/(3!)(x-2)^3 +...
=1-2/3(x-2) +(8/9)/2(x-2)^2 +(-16/9)/6(x-2)^3 +...
=1-2/3(x-2) +4/9(x-2)^2 +8/27(x-2)^3 +...
= sum_(n=0)^oo (-(2(x-2))/3)^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=0)^oo (-(2(x-2))/3)^n with  sum_(n=0)^oo a*r^n , we determine: r = -(2(x-2))/3 .
Apply the condition for convergence of geometric series: |r|lt1 .
|-(2(x-2))/3|lt1
|-1|*|(2(x-2))/3|lt1
1*|(2(x-2))/3|lt1
|(2(x-2))/3|lt1
|(2x-4)/3|lt1
-1lt(2x-4)/3lt1
Multiply each sides by 3 :
-1*3lt(2x-4)/3*3lt1*3
-3lt2x-4lt3
Add 4 on each sides:
-3+4lt2x-4+4lt3+4
1lt2xlt7
Divide each side by 2 :
1/2lt2x/2lt7/2
1/2ltxlt7/2
Thus, the power series  of the function f(x) =3/(2x-1) centered at c=2 is sum_(n=0)^oo (-(2(x-2))/3)^n  and has an interval of convergence: 1/2ltxlt7/2 .

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