Figuring out coefficients of composition of a first degree polynomial into a quadratic

Question

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

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.

Answer 1

A more easy way is the following :

If $f(x)=ax+b$, then $$g \circ f(x) = (ax+b)^2 + (ax+b) - 2 = a^2x^2 + (2ab+a)x + (b^2+b-2)$$

By identification, you must have $a^2 = 4$, $2ab+a = -10$ and $b^2+b-2 = 4$. If $a=2$, you obtain $b=-3$ which works ; if $a=-2$, you obtain $b=2$ which also works. So you have two solutions : $$f(x) = 2x-3 \quad \text{ and } \quad f(x)=-2x+2 $$

Answer 2

Your "completing the square" could also be used this way. We have $$ \ g(x) \ = \ x^2 + x - 2 \ = \ (x+2)·(x-1) \ = \ \left( x + \frac12 \right)^2 - \frac94 \ \ $$ and $$ \ [g \circ f](x) \ = \ 2·(2x^2 - 5x + 2) \ = \ 2·(2x-1)·(x-2) \ = \ 4· \left( x - \frac54 \right)^2 - \frac94 \ \ . $$ The zeroes of these polynomials are then $ \ -\frac12 \pm \frac32 \ = \ -2 \ , \ 1 \ \ $ and $ \ \frac54 \pm \frac{3}{2·\sqrt4} \ = \ \frac12 \ , \ 2 \ \ . $

The composition $ \ [g \circ f](x) \ = \ g(ax+b) \ \ $ corresponds to curve transformations of a "horizontal stretch" and "horizontal shift" for $ \ a > 0 \ $ and includes a "horizontal reflection" about the $ \ y-$axis for $ \ a < 0 \ \ . $ There is no effect on $ \ y-$coordinates of points.

The vertex of the parabola for $ \ g(x) \ $ is then located at $ \ \left(-\frac12 \ , \ -\frac94 \right) \ \ $ and is to be transformed to the vertex $ \ \left( \frac54 \ , \ -\frac94 \right) \ \ . $ There are two possibilities for the transformation of the zeroes:

• for $ \ a > 0 \ \ , $ the "stretch and shift" is itself a composition $ \ a·\left(x + \frac{b}{a} \right) \ ; $

• for $ \ a < 0 \ \ , $ the added "reflection" makes this $ \ -|a|·\left(x - \frac{b}{|a|} \right) \ . $

The transformations are thus $$ \left( \ [ax+b] \ + \ \frac12 \ \right)^2 \ \ = \ \ a^2·\left( \ x \ + \ \frac{b}{a} \ + \ \frac{1}{2·a} \ \right)^2 \ \ = \ \ 4· \left( x - \frac54 \right)^2 $$ $$ \Rightarrow \ \ a^2 \ = \ 4 \ \ \Rightarrow \ \ a \ = \ \pm 2 \ \ , $$ for which we have either $$ a \ = \ +2 \ \ \Rightarrow \ \ \frac{b}{2} \ + \ \frac{1}{2·2} \ \ = \ \ - \frac54 \ \ \Rightarrow \ \ b \ = \ -3 \ \ $$ or $$ a \ = \ -2 \ \ \Rightarrow \ \ \frac{b}{-2} \ + \ \frac{1}{2·[-2]} \ \ = \ \ - \frac54 \ \ \Rightarrow \ \ b \ = \ +2 \ \ . $$

We can check on how the zeroes are transformed to verify these results:

$$ \mathbf{g( 2x - 3 ) = 0 \ : } \quad 2x - 3 \ = \ -2 \ \ \rightarrow \ \ x \ = \ \frac{-2}{2} \ + \ \frac32 \ = \ \frac12 \ \ \ , $$ $$ 2x - 3 \ = \ 1 \ \ \rightarrow \ \ x \ = \ \frac12 \ + \ \frac32 \ = \ 2 \ \ , $$ so points on the parabola are effectively "compressed" toward the symmetry axis, and said axis translates "to the right";

$$ \mathbf{g( -2x + 2 ) = 0 \ : } \quad -2x + 2 \ = \ -2 \ \ \rightarrow \ \ x \ = \ \frac{-2}{-2} \ + \ \frac{2}{-2} \ = \ 2 \ \ \ , $$ $$ -2x + 2 \ = \ 1 \ \ \rightarrow \ \ x \ = \frac{1}{-2} \ + \ \frac{2}{-2} \ = \ \frac12 \ \ , $$ so points are "compressed" toward the symmetry axis, reflected around it, and effectively the axis translates "to the right".