F(x)= 2x-1/x+1, g(x)=x²-1, h(x)=3x+5. tentukan (g o h o f)^-1(x)

F(x)= 2x-1/x+1, g(x)=x²-1, h(x)=3x+5. tentukan (g o h o f)^-1(x)

Jawaban:

$f(x) = frac{2x - 1}{x + 1} \ y= frac{2x - 1}{x + 1} \ xy + y = 2x - 1 \ y + 1 = 2x - xy \ y + 1 = x(2 - y) \ x = frac{y + 1}{2 - y} \ {f}^{ - 1} (x) = frac{x + 1}{2 - x}$

$h(x) = 3x + 5 \ y = 3x + 5 \ 3x = y - 5 \ x = frac{y - 5}{3} \ {h}^{ -1 } (x) = frac{x - 5}{3}$

${(g : o : h : f)}^{ - 1} (x) = {g}^{ - 1} ( {h}^{ - 1} ( {f}^{ - 1}(x ))\ {h}^{ - 1} (x) = frac{x - 5}{3} \ {h}^{ - 1} ( {f}^{ - 1} (x))= frac{( frac{x + 1}{2 - x}) - 5 }{3} \ = frac{ frac{x + 1 - 5(2 - x)}{2 - x} }{3} \ = frac{x + 1 - 10 + 5x}{3(2 - x)} \ = frac{6x - 9}{6 - 3x}$

$g(x) = {x}^{2} - 1 \ y = {x}^{2} - 1 \ {x}^{2} = y + 1 \ x = sqrt{y + 1} \ {g}^{ - 1} (x) = sqrt{y + 1}$

${(g : o : h : o : f)}^{ - 1} (x) = {g}^{ - 1} ( {h}^{ - 1} ( {f}^{ - 1}(x ))\ {h}^{ - 1} (x) = frac{x - 5}{3} \ {h}^{ - 1} ( {f}^{ - 1} (x))= frac{( frac{x + 1}{2 - x}) - 5 }{3} \ = frac{ frac{x + 1 - 5(2 - x)}{2 - x} }{3} \ = frac{x + 1 - 10 + 5x}{3(2 - x)} \ = frac{6x - 9}{6 - 3x} \ {g}^{ - 1} (x) = sqrt{x + 1} \ {g}^{ - 1} ( {h}^{ - 1} ( {f}^{ - 1}(x )) = sqrt{ frac{6x - 9}{6 - 3x} + 1 } \ sqrt{ frac{3(2x - 3)}{3(2x - 1)} + 1} \ = sqrt{ frac{2x - 3}{2x - 1} + 1 } \$