In order to determine when the function is increasing or decresaing, we shall begin by taking its derivative, as follows;
[tex]\begin{gathered} f(x)=\frac{3x+7}{2x-5} \\ y^{\prime}=\frac{(2x-5)-(3x+7)}{(2x-5)^2} \\ We\text{ simplify and we now have;} \\ y^{\prime}=\frac{2x-5-3x-7}{(2x-5)^2} \\ y^{\prime}=\frac{-x-12}{(2x-5)^2} \end{gathered}[/tex]The numerator of the derivative is negative. When the result is a negative that means the function is decreasing.
Also, the inverse of the function f(x) is derived as follows;
[tex]\begin{gathered} f(x)=\frac{(3x+7)}{(2x-5)} \\ We\text{ re-write this as an equation as follows;} \\ y=\frac{3x+7}{2x-5} \\ We\text{ now switch places for variables x and y;} \\ x=\frac{3y+7}{2y-5} \\ \text{Cross multiply and you'll have;} \\ x(2y-5)=3y+7 \\ 2xy-5x=3y+7 \\ \text{Subtract 3y from both sides;} \\ 2xy-5x-3y=7 \\ \text{Add 5x to both sides;} \\ 2xy-3y=7+5x \\ \text{Factor out y from the left side;} \\ y(2x-3)=7+5x \\ \text{Divide both sides by }(2x-3) \\ y=\frac{(7+5x)}{(2x-3)} \\ We\text{ can now re-write this using function notation} \\ f^{-1}(x)=\frac{(7+5x)}{(2x-3)} \end{gathered}[/tex]ANSWER:
The function f(x) is decreasing at the point of discontinuity (when the function is undefined) which is
[tex]\begin{gathered} 2x-5=0 \\ 2x=5 \\ x=\frac{5}{2} \\ x=2.5 \end{gathered}[/tex]The inverse of the function is;
[tex]f^{-1}(x)=\frac{7+5x}{2x-3}[/tex]
