Finding Inverse of P(r) = 2r^2 + 2r - 1

[Don't Drink and Derive]

Took a picture of my work. At which step did I miscalculate? Used online inverse calculator to check my work and I am way off. Thanks!

 

[mmm4444bot]

Took a picture of my work. At which step did I miscalculate? Used online inverse calculator to check my work and I am way off. Thanks!

I noticed three issues, but I might have missed something while twisting my head. In the future, please rotate your images before posting them.

Your method of completing the square using both sides of an equation is okay, but don't set P to zero.

That way, your result at step 3 (after adding 1 to each side and dividing each side by 2) will be:

r^2 + r = (P + 1)/2

At step 4, it looks like you correctly added 1/4 to the left-hand side, but you forgot to add it to the right-hand side.

Once you have (r + 1/2)^2 = (P + 1)/2 + 1/4, you need to finish completing the square, by solving for P.

You set up finding the inverse correctly, by swapping symbols r and y, but remember the following rule.

sqrt[(P + 1/2)^2] = ±(P + 1/2)

That is, if you take the square root of a symbolic square, you need to consider both the positive and negative root.

Cheers :cool:

 

[mmm4444bot]

Here's a plot of P(r) and its inverse.

Note that the inverse (shown in red) comes in two pieces -- one from the positive root and one from the negative root.

If you forget to consider both roots, your answer for P^-1(r) will only be half correct.

I included the line y=x (in green), to show the symmetry about that line of a function and its inverse.

 

[Don't Drink and Derive]

Thanks for the help!