1. ## function inverse

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.

2. Hello,

$\displaystyle x=y^2-8y$

And you want y with respect to x.

$\displaystyle y^2-8y-x=0$

Solve this quadratic equation, treating x as a constant.

Be aware that the range of x is particular, because if the discriminant is negative, there is no real solution.

3. ## Tricky

$\displaystyle y=x^2-8x \Rightarrow x^2-8x-y=0$ know how to solve for x now?

Spoiler:

$\displaystyle x=\frac{8\pm \sqrt{64+4y}}{2}=\frac{8\pm \sqrt{4(16+y}}{2}=4\pm \sqrt{16+y}$

Just need a function to work as the inverse, so might as well take the positive one and check it to make sure it works. The inverse is $\displaystyle f^{-1}(x)=4+\sqrt{16+x}$.

$\displaystyle f(4+\sqrt{16+x})=(4+\sqrt{16+x})^2-8(4+\sqrt{16+x})$
$\displaystyle =16+8\sqrt{16+x}+16+x-32-8\sqrt{16+x}=x$

Check to make sure it is a two sided inverse
$\displaystyle f^{-1}(f(x))=4+\sqrt{16+(x^2-8x)}=4+\sqrt{(x-4)^2}=4+(x-4)=x$

4. i forgot to mention that it stated x<=4 (for f(x))

5. Originally Posted by furor celtica
i forgot to mention that it stated x<=4 (for f(x))
A very good thing to have said initially because the function you gave does not have an inverse! Solving the equation y= f(x) for x gives two different solutions depending on whether you take the positive or negative square root. But those two solutions "meet" at x= 4. That is, one sign gives x>4, the other <4.

6. Originally Posted by Moo
Hello,

$\displaystyle x=y^2-8y$

And you want y with respect to x.

$\displaystyle y^2-8y-x=0$

Solve this quadratic equation, treating x as a constant.

Be aware that the range of x is particular, because if the discriminant is negative, there is no real solution.

Can some1 explain to me how to solve this equation? As it seems there is 2 constants?

7. Originally Posted by furor celtica
f(x) = x^2 - 8x, find f^-1 (x)
... i forgot to mention that it stated x<=4 (for f(x))
$\displaystyle y = x^2-8x$

$\displaystyle x = y^2-8y$

use the method of completing the square ...

$\displaystyle x+16 = y^2-8y+16$

$\displaystyle x+16 = (y - 4)^2$

$\displaystyle \pm \sqrt{x+16} = y-4$

$\displaystyle y = 4 \pm \sqrt{x+16}$

since $\displaystyle x \le 4$ for $\displaystyle f(x)$ , $\displaystyle f^{-1}(x) \le 4$ ...

$\displaystyle f^{-1}(x) = 4 - \sqrt{x+16}$