Answer:
[tex]g^{-1}(x)=-1\pm\frac{\sqrt{x(x+3)}}{x}[/tex]
Step-by-step explanation:
Given : [tex]g(x)= \frac{3}{x^2+2x}[/tex]
To find : [tex]g^{-1}(x)[/tex]
Solution :
We have given the function,
[tex]g(x)= \frac{3}{x^2+2x}[/tex]
To find inverse let y=g(x)
[tex]y= \frac{3}{x^2+2x}[/tex]
Replace the value of x and y,
[tex]x=\frac{3}{y^2+2y}[/tex]
Solve for y,
[tex]x(y^2+2y)=3[/tex]
[tex]xy^2+2xy-3=0[/tex]
Solve by quadratic formula,
i.e. The equation [tex]ax^2+bx+c=0[/tex] has solution
[tex]x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}[/tex]
On comparing, a=x , b=2x , c=-3
[tex]y=\frac{-(2x)\pm\sqrt{(2x)^2-4(x)(-3)}}{2(x)}[/tex]
[tex]y=\frac{-2x\pm\sqrt{4x^2+12x}}{2x}[/tex]
[tex]y=\frac{-2x\pm2\sqrt{x(x+3)}}{2x}[/tex]
[tex]y=-1\pm\frac{\sqrt{x(x+3)}}{x}[/tex]
Therefore, The inverse of g(x) is
[tex]g^{-1}(x)=-1\pm\frac{\sqrt{x(x+3)}}{x}[/tex]
