If g(x) = x^2 + x - 2 and frac{1}{2}g(f(x)) = | vedclass

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(A) Given $\frac{1}{2}g(f(x)) = 2x^2 - 5x + 2$,we have $g(f(x)) = 4x^2 - 10x + 4$.Since $g(x) = x^2 + x - 2$,substituting $f(x)$ into $g(x)$ gives:$(f

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$2x - 3$

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(A) Given $\frac{1}{2}g(f(x)) = 2x^2 - 5x + 2$,we have $g(f(x)) = 4x^2 - 10x + 4$.Since $g(x) = x^2 + x - 2$,substituting $f(x)$ into $g(x)$ gives:$(f(x))^2 + f(x) - 2 = 4x^2 - 10x + 4$.Rearranging the terms,we get $(f(x))^2 + f(x) - (4x^2 - 10x + 6) = 0$.This is a quadratic equation in $f(x)$. Using the quadratic formula $f(x) = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$:$f(x) = \frac{-1 \pm \sqrt{1^2 - 4(1)(-(4x^2 - 10x + 6))}}{2(1)}$$f(x) = \frac{-1 \pm \sqrt{1 + 16x^2 - 40x + 24}}{2}$$f(x) = \frac{-1 \pm \sqrt{16x^2 - 40x + 25}}{2}$$f(x) = \frac{-1 \pm \sqrt{(4x - 5)^2}}{2}$$f(x) = \frac{-1 \pm (4x - 5)}{2}$.Taking the positive root,$f(x) = \frac{-1 + 4x - 5}{2} = \frac{4x - 6}{2} = 2x - 3$.

If $g(x) = x^2 + x - 2$ and $\frac{1}{2}g(f(x)) = 2x^2 - 5x + 2$,then $f(x)$ is

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