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The question is to find the inverse of the following function -  p(r) = 2r^2+2r-1

I realise that this cannot be factored by decomposition. So I'm using the quadratic formula. I've done most of the question but stuck near the end.  I have noted the final answer on the side but i dont know how to get it to it. Can you help to finish this question? 

UpTheChels  Sep 15, 2017
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3+0 Answers

 #1
avatar+5249 
+2

I followed your work..I think you've got it right so far!

(Except  c = (-1 - y) , but you had it right in the problem.)   smiley
 

You are very very close to the answer. Maybe this will nudge you in the right direction.

 

\(x=\frac{-2\,\pm\,2\sqrt{2y\,+\,3}}{4} \\~\\ x=\frac{2(-1\,\pm\,\sqrt{2y\,+\,3})}{2(2)}\)    Factor a 2 out of the numerator and factor a 2 out of the denominator.

 

Below is the rest of the answer if you need more help.

 

 

 

 

 

 

----------

 

 

 

\(x=\frac{2(-1\,\pm\,\sqrt{2y\,+\,3})}{2(2)}\)

                                                Divide the numerator and denominator by  2  .

\(x=\frac{-1\,\pm\,\sqrt{2y\,+\,3}}{2}\)

                                                Rearrange the terms in the numerator.

\(x=\frac{\pm\,\sqrt{2y\,+\,3}\,-\,1}{2} \\~\\ \begin{array}\ x=\frac{\sqrt{2y\,+\,3}\,-\,1}{2} \qquad\text{ or }\qquad &x=\frac{-\,\sqrt{2y\,+\,3}\,-\,1}{2}& \\~\\ & x=\frac{-1(\,\sqrt{2y\,+\,3}\,+\,1)}{2}&\\~\\ & x=-\,\frac{\,\sqrt{2y\,+\,3}\,+\,1}{2}& \end{array}\)

 

Also...for your final answer, do you need to say  p-1(r)  for  x  and  r  for  y  ?

hectictar  Sep 15, 2017
edited by hectictar  Sep 15, 2017
 #2
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+1

Thank you so much, that makes a lot of sense. For the final answer, I was wondering what exactly it would be myself. Since the original question is p(r), it only makes sense that I write it as p^-1 (r).  

Guest Sep 15, 2017
 #3
avatar+78719 
+2

Thanks, hectictar.......here's another possible approach done by completing the square

 

Let's write "y"  for  p(r).......we'll come back to this later

 

y  = 2r^2 + 2r -1        add 1 to both sides

 

y + 1   =  2r^2 + 2r

 

y + 1  =  2 [ r^2 + r]     divide both sides by 2

 

[ y + 1] / 2    =  r^2 + r        completet the square on r

 

[ y + 1 ] / 2  + 1/4  +     =    r^2 +  r  + 1/4     simplify

 

[ 2y + 2 + 1 ] / 4   =   ( r + 1/2)^2

 

[ 2y + 3 ]  / 4   =   ( r + 1/2)^2       take both roots

 

±  √ [2y + 3]  / 2  =   r  + 1/2

 

±  √ [2y + 3]  / 2   - 1/2   =  r

 

± ( √ [2y + 3]    - 1  )  / 2   = r          "exchange"  r and y

 

± ( √ [2r + 3]    - 1  )  / 2   =  y

 

For "y"  write  p-1(r)

 

p-1 (r)  =   ± ( √ [2r + 3]    - 1  )  / 2 

 

 

 

cool cool cool

CPhill  Sep 15, 2017

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