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`(-oo,5]``[1,oo)``(-oo,1) cup [5, oo)`None of these

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`f(x)+f((x-1)/(x))=1+x " (1)" ` <br> In (1), replace x by `(x-1)/(x)`. Then <br> `f((x-1)/(x))+f(((x-1)/(x)-1)/((x-1)/(x)))=1+(x-1)/(x)` <br> `or f((x-1)/(x))+f((1)/(1-x))=1+(x-1)/(x) " (2)" ` <br> Now, from `(1) -(2)`, we have <br> `f(x)-f((1)/(1-x))=x-(x-1)/(x) " (3)" ` <br> In (3), replace x by `(1)/(x-1)`. Then <br> `f((1)/(1-x))-f((x-1)/(x))=(1)/(1-x)-((1)/(1-x)-1)/((1)/(1-x))` <br> ` or f((1)/(1-x))-f((x-1)/(x))=(1)/(1-x)-x " (4)" ` <br> Now, from `(1)+(3)+(4)`, we have <br> `2f(x)=1+x+x-(x-1)/(x)+(1)/(1-x)-x` <br> ` or f(x)=(x^(3)-x^(2)-1)/(2x(x-1))` <br> `y=g(x)=(x^(2)-x-1)/(x(x-1)) or (y-1)x^(2)+(1-y)x+1=0` <br> Now, x is real, Therefore, ` D ge 0 or (1-y)^(2)-4(y-1) ge 0` <br> `or (y-1) (y-5) ge 0` <br> `or y in (-oo,1) cup [5,oo)`**Introduction: average rate and instantaneous rate**

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