jika f(x)=2x + 7 dan (g •f )(x)=4x²+20x+5 tentukan a.g(x) b.g^-1(x)

[tex]f(x) = 2x + 7 \: = > \: {f}^{ - 1}(x) = \frac{x - 7}{2} \\ (gof)(x) = 4 {x}^{2} + 20x + 5 \\ g(x) = 4( { \frac{x - 7}{2} )}^{2} + 20( \frac{x - 7}{2} ) + 5 \\ = 4( \frac{ {x}^{2} - 14x + 49 }{4} ) + 2( \frac{10x - 70}{2} ) + 5 \\g(x) = {x }^{2} - 4x - 16 \\ g {}^{ } (x) = y \\ {x}^{2} - 4x - 16 = y \\ (x - 2) {}^{2} - 32 = y \\ {(x - 2)}^{2} = y + 32 \\ x - 2 = \sqrt{y + 32} \\ x = \sqrt{y + 32} + 2 \\ {g}^{ - 1} (x)= \sqrt{x + 32} + 2[/tex]

(g • f)(x) = 4x²+20x+5
(g(f(x)) = 4x²+20x+5
g(2x+7) = 4x²+20x+5
[tex]g(x) = 4 (\frac{x-7}{2})^{2} + 20( \frac{x-7}{2}) + 5 \\ \\ g(x) = x^{2} -14x+49 + 10x-70 + 5 \\ \\ g(x) = x^{2} -4x-16[/tex]

y = x²-4x-16
y = (x²-4x+4)-20
y = (x-2)²-20
y+20 = (x-2)²
[tex] \sqrt{y+20} = x-2 \\ x = \sqrt{y+20} +2 \\ g^{-1}(x) = \sqrt{x+20} +2[/tex]