Functions

To find the domain of the real-valued function $f(x)=\;\frac{1}{\sqrt{\log_{0.5}(2x-3)}}+\sqrt{4-9x^2}$, we analyze the conditions under which each part of the function is defined.First Part: $\;\frac{1}{\sqrt{\log_{0.5}(2x-3)}}$For the square root in the denominator to be defined and non-zero, the expression inside must be positive: $\log_{0.5}(2x-3)>0$.The base of the logarithm is less than 1 . This means:$2x-3 < 1 \;\; \Rightarrow \;\; x < 2$Additionally, the expression inside the logarithm must be positive:$2x-3 > 0 \;\; \Rightarrow \;\; x > \frac{3}{2}$Second Part: $\sqrt{4-9x^2}$The expression under the square root must be non-negative:$4-9x^2 > 0 \;\; \Rightarrow \;\; x^2 < \frac{4}{9}$Solving for $x$, we derive:$-\frac{2}{3} < x < \frac{2}{3}$Intersection of Conditions:From the first part, we need $x$ to satisfy both conditions: $\frac{3}{2} < x < 2$.From the second part, $x$ must be in the interval $-\frac{2}{3} < x < \frac{2}{3}$.The intersection of these conditions is empty, as there is no number that simultaneously satisfies $\frac{3}{2} < x < 2$ and $-\frac{2}{3} < x < \frac{2}{3}$.Hence, the domain of the function is a null set.