> 
> given $g(x)=x^2 + x-2$ and $ g \circ f = 2 [2x^2-5x +2]$, find $f(x)$ ( $f(x)$ is form $ax+b$)
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> 
> 

I found the inverse of g as two functions,

$$ y =( x + \frac12)^2 - \frac94 $$

$$ \pm (\sqrt{y + \frac94} -\frac12 ) = x$$

or,

$ g^{-1} (y) = \pm (\sqrt{y + \frac94} -\frac12 ) $

Ok, so into this function I plugged in $ y= g \circ f = 2 [2x^2 - 5x +2]$

and this gave me:

$ f(x) = \begin{cases} 2x -3 \\
 2-2x \\
 \end{cases}$

Now there are two different functions which f(x) could have been.

Now, If I try do this another way, that is, using derivatives,

$$ g( ax+b) = (ax+b)^2 + (ax+b) -2$$

$$ g \circ f = 2 [ 2x^2 - 5x +2]$$

Equating the two different definitions,

$$ 2 [ 2x^2 - 5x +2]=(ax+b)^2 + (ax+b) -2$$

Taking a derivative,

$$ 2 ( ax+b)a + a- 2 = 8x - 10$$

And one more,

$$ 2 a^2 = 8 $$

$$ a^2= 4$$

Evaluating the equation after first derivative at x=0, I get $ b=-6$

so, $ f(x) = 4x - 6$

which is inconsistent with the previous answer. What went wrong?

Edit: fixed a method in derivative.