I'm guessing the function is
[tex]f(x)=\dfrac8{x^2-4x-12}=\dfrac8{(x-6)(x+2)}[/tex]
which, split into partial fractions, is equivalent to
[tex]\dfrac1{x-6}-\dfrac1{x+2}[/tex]
Recall that for [tex]|x|<1[/tex] we have
[tex]\dfrac1{1-x}=\displaystyle\sum_{n=0}^\infty x^n[/tex]
With some rearranging, we find
[tex]\dfrac1{x-6}=-\dfrac16\dfrac1{1-\frac x6}=\displaystyle-\frac16\sum_{n=0}^\infty\left(\frac x6\right)^n[/tex]
valid for [tex]\left|\dfrac x6\right|<1\[/tex], or [tex]|x|<6[/tex], and
[tex]\dfrac1{x+2}=\dfrac12\dfrac1{1-\left(-\frac x2\right)}=\displaystyle\frac12\sum_{n=0}^\infty\left(-\frac x2\right)^n[/tex]
valid for [tex]\left|-\dfrac x2\right|<1[/tex], or [tex]|x|<2[/tex].
So we have
[tex]f(x)=\displaystyle-\frac16\sum_{n=0}^\infty\left(\frac x6\right)^n-\frac12\sum_{n=0}^\infty\left(-\frac x2\right)^n[/tex]
[tex]f(x)=\displaystyle-\sum_{n=0}^\infty\left(\frac{x^n}{6^{n+1}}+\frac{(-x)^n}{2^{n+1}}\right)[/tex]
[tex]f(x)=\displaystyle-\sum_{n=0}^\infty\frac{1+3(-3)^n}{6^{n+1}}x^n[/tex]
Taken together, the power series for [tex]f(x)[/tex] can only converge for [tex]|x|<2[/tex], or [tex]-2<x<2[/tex].
The interval of convergence can only occur at |x|<2 or -2<x<2
Given that the function is:
[tex]f(x) = 8/x^{2} -4x-12= 8/(x-6)(x+2)[/tex]
When this is split into partial fractions, we would get
1/x-6 - 1/ x+2
When this is rearranged, it would be:
[tex]1/x-6 = 1/6 .1/1-x/6= -1/6[/tex]Σ to infinity, n=0 (x/6)^n
Taken together, the power series for f(x) can only converge for |x|<2 or -2<x<2.
Read more about the convergence of series here:
https://brainly.com/question/2254113
