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Suppose that the inverse of the function \(f(x) = \sqrt{ax + b}\) is given by \(f^{-1}(x) = -(a+2)x^2+(2b+1)\) for all \(x\)satisfying \(ax+b \ge 0\). Find \(a+b\).

 

Much Help would be appreciated! 

 Oct 29, 2019
 #1
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y  = sqrt  (ax + b)          square both sides

 

y^2  =  ax + b             subtract b from both sides

 

y^2 - b  =  ax            divide both sides by  a

 

y^2 / a  - b/a   =    x             "swap"  x  and y

 

x^2 / a  -  b/a   =   the inverse

 

So.....we have that

 

(1/a)x^2  -  b/a   =  -(a + 2)x^2  +  (2b + 1)

 

This must mean that

 

1/a   = -(a + 2)

 

1/a  = -a - 2

 

a + 1/a  + 2  = 0        multiply through by a    and rearrange

 

a^2 + 2a  + 1  =  0     factor

 

(a + 1)^2  =  0            take the root

 

a + 1   =   0

 

a  = -1

 

So.....we also have that

 

-b/a  = 2b + 1

 

-b/ (-1)  =  2b + 1

 

b  = 2b + 1

 

-1   = b

 

So  

 

a + b  = 

 

-1  +  -1   =

 

-2

 

cool cool cool

 Oct 29, 2019
 #2
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+1

thank you so much!!!!

 Oct 29, 2019

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