Given:
The expression is,
[tex]\frac{1}{2}+\frac{15}{2x-14}-\frac{3x-15}{x^2-7x}[/tex]
Explanation:
Simplify the expression.
[tex]\begin{gathered} \frac{1}{2}+\frac{15}{2x-14}-\frac{3x-15}{x^2-7x}=\frac{1}{2}+\frac{15}{2(x-7)}-\frac{3x-15}{x^{}(x-7)} \\ =\frac{x(x-7)+15\cdot x-(3x-15)\cdot2}{2x(x-7)} \\ =\frac{x^2-7x+15x-6x+30}{2x(x-7)} \\ =\frac{x^2+2x+30}{2x(x-7)} \end{gathered}[/tex]
The function is not defined at which denominator is equal to 0. So,
[tex]\begin{gathered} 2x(x-7)=0 \\ x=0,7 \end{gathered}[/tex]
Thus function is defined for all values of x except 0 and 7. So domain of the function is,
[tex](-\infty,0)\cup(0,7)\cup(7,\infty)[/tex]
