What is the sum of extremes of the function f(x)=(x^2+3x-3)/(x-1) ?

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We have to find the sum of the extremes of f(x) = (x^2 + 3x - 3) / (x-1).

We have to differentiate f(x).

f'(x) = [(x^2 + 3x - 3)'*(x - 1) - (x^2 + 3x - 3)*(x - 1)']/(x - 1)^2

=> [(2x + 3)(x - 1) - (x^2 + 3x - 3)] /( x- 1)^2

=> [2x^2 + 3x - 2x - 3 - x^2 - 3x + 3]/ ( x - 1)^2

=> [x^2 - 2x]]/(x - 1)^2

=> (x^2 - 2x)/(x - 1)

Equating this to zero, we get x = 0 and x = 2

At x = 0 the extreme value of the function is 3.

At x = 2 the extreme value of the function is 2^2 + 3*2 - 3 = 7

The required sum is 10.

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