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Jawaban Matematika hal.114

6.
$f^{-1}(x)=\frac{x-1}{5}\\f^{-1}(y)=\frac{y-1}{5}\\x=\frac{y-1}{5}\\5x=y-1\\5x+1=y\\5x+1=f(x)\\\\g^{-1}(x)=\frac{3-x}{2}\\g^{-1}(y)=\frac{3-y}{2}\\x=\frac{3-y}{2}\\2x=3-y\\y=3-2x\\g(x)=3-2x$

$(f\circ g)(x)=f(g(x))\\(f\circ g)(x)=5(3-2x)+1\\(f\circ g)(x)=15-10x+1\\(f\circ g)(x)=16-10x\\\\y=16-10x\\10x=16-y\\x=\frac{16-y}{10}\\(f\circ g)^{-1}(y)=\frac{16-y}{10}\\\boxed{(f\circ g)^{-1}(x)=\frac{16-x}{10}}$

$(f\circ g)(x+1)=-2x^2-4x-1\\f(g(x+1))=-2x^2-4x-1\\2(g(x+1))+1=-2x^2-4x-1\\2(g(x+1))=-2x^2-4x-1-1\\2(g(x+1))=-2x^2-4x-2\\g(x+1)=\frac{-2x^2-4x-2}{2}\\g(x+1)=\frac{-2(x^2+2x+1)}{2}\\g(x+1)=-(x^2+2x+1)\\g(x+1)=-(x+1)^2\\g(x)=-(x)^2\\g(x)=-x^2$

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

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Senin, 22 Januari 2018

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