Answer
No, it does not converge at $x=2$.
Work Step by Step
By Theorem 3:
${\left( {1 + x} \right)^a} = 1 + \dfrac{a}{{1!}}x + \dfrac{{a\left( {a - 1} \right)}}{{2!}}{x^2} + \dfrac{{a\left( {a - 1} \right)\left( {a - 2} \right)}}{{3!}}{x^3} + \cdot\cdot\cdot + \left( {\begin{array}{*{20}{c}}
a\\
n
\end{array}} \right){x^n} + \cdot\cdot\cdot$
converges for $\left| x \right| \lt 1$.
Setting $a = \dfrac{3}{4}$ in the series above gives
$f\left( x \right) = {\left( {1 + x} \right)^{3/4}} = 1 + \dfrac{{\dfrac{3}{4}}}{{1!}}x + \dfrac{{\dfrac{3}{4}\left( { - \dfrac{1}{4}} \right)}}{{2!}}{x^2} + \dfrac{{\dfrac{3}{4}\left( { - \dfrac{1}{4}} \right)\left( { - \dfrac{5}{4}} \right)}}{{3!}}{x^3} + \cdot\cdot\cdot + \left( {\begin{array}{*{20}{c}}
{3/4}\\
n
\end{array}} \right){x^n} + \cdot\cdot\cdot$
Since the series converges for $\left| x \right| \lt 1$, it does not converge at $x=2$.
At $x=2$, the terms are increasing and the series approaches infinity, therefore it does not converge. Moreover, since the term does not converges to zero. By the $n$th Term Divergence Test, the series diverges at $x=2$.
