function inverse

[red and white kop!]

f(x) = x^2 - 8x
find f^-1 (x)
i know this must be very simple but i have no idea what to do here. can somebody just show me how it is done? i am really at a loss of what to do, especially since i usually have no problem with inverse functions. i just use f(x)= y and solve for x, then replace y with x.

 

[DrMike]

f(x) = x^2 - 8x
find f^-1 (x)
i know this must be very simple but i have no idea what to do here. can somebody just show me how it is done? i am really at a loss of what to do, especially since i usually have no problem with inverse functions. i just use f(x)= y and solve for x, then replace y with x.

That's right. Folowing your steps, you get : y = x[sup:31ysa5fh]2[/sup:31ysa5fh]-8x, that is, x[sup:31ysa5fh]2[/sup:31ysa5fh]-8x-y=0.

You need to use the quadratic formula to solve this :

$\displaystyle ax^2+bx+c=0$ gives $\displaystyle x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}$.

Here, that means

$\displaystyle x=\frac{8\pm\sqrt{64+4y}}{2}$

Exchanging x and y gives :

$\displaystyle y=\frac{8\pm\sqrt{64+4x}}{2}$

Going back to $\displaystyle x=\frac{8\pm\sqrt{64+4y}}{2}$, this means deciding whether your original function f(x) is defined on a sub-interval of $\displaystyle x\geq4$ or of $\displaystyle x\leq 4$. If it's defined everywhere, it doesn't have an inverse function.

 

[red and white kop!]

thanks, i appreciate