If g(x) = x^2 + x − 2 and (1/2) gof(x) = 2x^2 − 5x + 2, then prove that f(x) = 2x − 3.


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Question Number 13598 by Tinkutara last updated on 21/May/17

$If g (x) = x^{2} + x - 2 and$ $\frac{1}{2} g o f (x) = 2 x^{2} - 5 x + 2, then prove$ $that f (x) = 2 x - 3 .$
Commented by mrW1 last updated on 21/May/17

$\frac{1}{2} g o f = \frac{1}{2} (f^{2} + f - 2) = 2 x^{2} - 5 x + 2$ $f^{2} + f - 2 (2 x^{2} - 5 x + 3) = 0$ $f = \frac{- 1 \pm \sqrt{1 + 4 \times 2 (2 x^{2} - 5 x + 3)}}{2}$ $f = \frac{- 1 \pm \sqrt{16 x^{2} - 40 x + 25}}{2}$ $f (x) = \frac{- 1 \pm (4 x - 5)}{2} = \frac{4 x - 6}{2} = 2 x - 3 o r \frac{- 4 x + 4}{2} = 2 (1 - x)$ $t h a t m e a n s t h e r e a r e 2 s o l u t i o n s :$ $f (x) = 2 x - 3$ $o r f (x) = 2 (1 - x)$
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