Answer :
Firstly remember that an inverse may not always exist.
If it does, then I find the following method the simplest:
1. Rewrite the equation [tex]f(x)=5x^2-1[/tex] as [tex]y=5x^2-1[/tex]
2. Swap the x and y's to get: [tex]x=5y^2-1[/tex]
3. Solve for y:
[tex]x=5y^2-1[/tex]
[tex]x+1=5y^2[/tex]
[tex]\frac{x+1}{5} = y^2[/tex]
[tex]\sqrt{\frac{x+1}{5}} =y[/tex]
[tex]\therefore y=\sqrt{\frac{x+1}{5}}[/tex]
This last step represents the inverse function
If it does, then I find the following method the simplest:
1. Rewrite the equation [tex]f(x)=5x^2-1[/tex] as [tex]y=5x^2-1[/tex]
2. Swap the x and y's to get: [tex]x=5y^2-1[/tex]
3. Solve for y:
[tex]x=5y^2-1[/tex]
[tex]x+1=5y^2[/tex]
[tex]\frac{x+1}{5} = y^2[/tex]
[tex]\sqrt{\frac{x+1}{5}} =y[/tex]
[tex]\therefore y=\sqrt{\frac{x+1}{5}}[/tex]
This last step represents the inverse function