Find the inverse of (x^2 - 4)/ 2x^2

Question

anonymous · Answer

Step one: Replace f(x) or what have you with y.

$$f(x)=\frac{x^2-4}{2x^2}$$ becomes $$y = \frac{x^2-4}{2x^2}$$

Step two: Swap x and y.
$$x=\frac{y^2-4}{2y^2}$$

Step three: Re-isolate y. I'll start you off by multiplying by 2y^2:
$$2y^2x=y^2-4$$

anonymous · Answer

Meh, I'll just do it now:
$$2y^2x-y^2=-4$$ We can factor y^2 out of the left hand side, like so: $$y^2(2x-1)=-4$$ Now, we can divide by the new quantity on the left. $$y^2 = \frac{-4}{2x-1}$$ Finally, to eliminate the exponent on the left hand side, we take the square root of both sides. Remember your plus/minus sign!
$$f^{-1}(x)=\pm\sqrt{\frac{-4}{2x-1}}$$

anonymous · Answer

https://www.desmos.com/calculator/2srcu4irgc Black = original Red = inverse Orange dashed = y=x (the line all inverse functions are reflected over!

artemisxrpg · Answer

Of course, I had to factor out the y^2. Thanks for the help.