Question

Find the inverses of the following equations

Answer 1

Where's the equation?

Answer 2

One sec

Answer 3

p(r)=2r^2+2r-1

Answer 4

That's my first equation whose inverse needs to be found algebraically

Answer 5

Hello

Answer 6

hi

Answer 7

do you know what it means for a function to be 1-to-1?

Answer 8

Yeah I do

Answer 9

Would you show me step by step you how did it?

Answer 10

I haven't actually done it. I can walk you through it. The function you put up above is quadratic, so it isn't one-to-one, so it's inverse won't be a function. To find the inverse algebraically, start by switching the p and r.

\[r = 2p^2+2p-1\]

Now solve for p by completing the square.

Answer 11

Ah ok basically to find an inverse function I switch the variables and then solve for the p

Answer 12

Completing the square.. I remember doing it a few years ago in high school but I forgot how to do that.

Answer 13

yes

Answer 14

\[r + 1=2(p^2+p)\]
\[\frac{ 1 }{ 2 }(r+1)=p^2+p\]
Then add (b/2)² to both sides
\[\frac{ 1 }{ 2 }(r+1)+\frac{ 1 }{ 4 }=p^2+p+\frac{ 1 }{ 4 }\]
This allows the right side to factor to
\[\frac{ 1}{2}r+\frac{ 3 }{ 4 }=\left( p+\frac{ 1 }{ 2 } \right)^2\]

Answer 15

Then when you take the square root, you'll get two different functions, each for a part of p(r)'s domain.

\[\pm\sqrt{\frac{ 1}{2}r+\frac{ 3 }{ 4 }}= p+\frac{ 1 }{ 2 }\]
\[p=-\frac{ 1 }{ 2 }\pm\sqrt{\frac{ 1}{2}r+\frac{ 3 }{ 4 }}\]