The inverse function h-1 is efined by h-1:x-2/3-x ,x does not equal to 3 , find h(x)
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Let y=h-1(x)=(x-2)/(3-x),

y(3-x)=x-2,

3y-xy=x-2,

x+xy=3y+2,

x(1+y)=3y+2, x=(3y+2)/(y+1).

But x=h(y), so h(y)=(3y+2)/(y+1) and h(x)=(3x+2)/(x+1).

by Top Rated User (981k points)

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