1. ## Inverse Function

I got to calculate the inverse function of a normal function.

The function is given by: Y=x-0.5x^2.
So I need the function which ginve x=..........
I am in doubt if there is an inverse function of this function.
I hope someone could provide me the correct answer.

2. Originally Posted by Divius
I got to calculate the inverse function of a normal function.

The function is given by: Y=x-0.5x^2.
So I need the function which ginve x=..........
I am in doubt if there is an inverse function of this function.
I hope someone could provide me the correct answer.

So you have to solve $\displaystyle x = y - \frac{y^2}{2}$ for y.

Option 1: Complete the square.

$\displaystyle x = - \frac{1}{2} \left( y^2 - 2y \right) = - \frac{1}{2} \left( [y - 1]^2 - 1 \right)$

$\displaystyle \Rightarrow -2x = [y - 1]^2 - 1 \Rightarrow 1 - 2x = [y - 1]^2$

$\displaystyle \Rightarrow \pm \sqrt{1 - 2x} = y - 1$

$\displaystyle \Rightarrow y = 1 \pm \sqrt{1 - 2x}$.

Option 2: Use the quadratic formula.

$\displaystyle 2x = 2y - y^2 \Rightarrow y^2 - 2y + 2x = 0$.

a = 1, b = -2, c = 2x:

$\displaystyle y = \frac{2 \pm \sqrt{4 - 8x}}{2} = \frac{2 \pm 2 \sqrt{1 - 2x}}{2} = 1 \pm \sqrt{1 - 2x}$.

In each case, the domain of $\displaystyle y = x - \frac{x^2}{2}$ will determine the range of the inverse and hence which of the $\displaystyle \pm \sqrt{1 - 2x}$ solutions is wanted.

3. Thank you very much for the clear answer to my question!
However, I calculated a mean value for y by repeating the
$\displaystyle y = x - \frac{x^2}{2}$

3 times for different x values. Next I divided the sum of the values by 3 to determine the mean y value. (in my example the mean y value is 0.007667)

I need to find the x value which is equal to this mean y value. So given a value of y I have to determine which x value corresponds with this value.

4. Originally Posted by Divius
Thank you very much for the clear answer to my question!
However, I calculated a mean value for y by repeating the
$\displaystyle y = x - \frac{x^2}{2}$

3 times for different x values. Next I divided the sum of the values by 3 to determine the mean y value. (in my example the mean y value is 0.007667)

I need to find the x value which is equal to this mean y value. So given a value of y I have to determine which x value corresponds with this value.
So you want to solve $\displaystyle 0.007667 = x - \frac{x^2}{2}$.

Re-arrange into a quadratic = 0.

Or just substitute y = 0.007667 into $\displaystyle x = 1 \pm \sqrt{1 - 2y}$.

5. This is what I was looking for, sorry I didn't used your first post as it makes it clear. However, I don't understand this step:
$\displaystyle y = \frac{2 \pm \sqrt{4 - 8x}}{2} = \frac{2 \pm 2 \sqrt{1 - 2x}}{2}$