If g(x) = x^2 + x - 2 and (g circ f)(x) = 2x^ | vedclass

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(A) Given $g(x) = x^2 + x - 2$ and $(g \circ f)(x) = g(f(x)) = 2x^2 - 5x + 2$.Let $f(x) = y$. Then $g(y) = y^2 + y - 2 = 2x^2 - 5x + 2$.$y^2 + y - 2 -

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$2x - 3$

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(A) Given $g(x) = x^2 + x - 2$ and $(g \circ f)(x) = g(f(x)) = 2x^2 - 5x + 2$.Let $f(x) = y$. Then $g(y) = y^2 + y - 2 = 2x^2 - 5x + 2$.$y^2 + y - 2 - (2x^2 - 5x + 2) = 0$.$y^2 + y - (2x^2 - 5x + 4) = 0$.Using the quadratic formula $y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$ where $a=1, b=1, c=-(2x^2 - 5x + 4)$:$y = \frac{-1 \pm \sqrt{1^2 - 4(1)(-(2x^2 - 5x + 4))}}{2}$.$y = \frac{-1 \pm \sqrt{1 + 8x^2 - 20x + 16}}{2} = \frac{-1 \pm \sqrt{8x^2 - 20x + 17}}{2}$.Since the options provided do not match this result, we test $f(x) = 2x - 3$ in $g(f(x))$:$g(2x - 3) = (2x - 3)^2 + (2x - 3) - 2 = (4x^2 - 12x + 9) + 2x - 3 - 2 = 4x^2 - 10x + 4 
eq 2x^2 - 5x + 2$.Testing $f(x) = 2x - 3$ was incorrect. Let's re-evaluate $g(f(x)) = (f(x)+2)(f(x)-1) = 2x^2 - 5x + 2 = (2x-1)(x-2)$.This implies $f(x) = 2x-1$ or $f(x) = x-2$. None of the options match. The question options are incorrect.

If $g(x) = x^2 + x - 2$ and $(g \circ f)(x) = 2x^2 - 5x + 2$, then $f(x)$ is equal to

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