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

To determine the local extremes of the function, we'll
have to determine the critical points first.

For this
reason, we'll have to determine the first derivative of f(x). Since the function is a
ratio, we'll apply the quotient rule:

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

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

We'll remove
the brackets:

f'(x) = (2x^2 + x -1)/(x -
1)^2

We'll determine the roots of the
numerator:

2x^2 + x -1 = 0

x1
= [-1+sqrt(1+8)]/4

x1 =
(-1+3)/4

x1 = 1/2

x2 =
-1

2x^2 + x -1 = (x + 1)(x -
1/2)

The critical values of the function are roots of the
1st derivative.

Now, we'll determine the local
extremes:

f(1/2) = (1/4 + 3/2 - 3)/(1/2 -
1)

f(1/2) =
(-5/4)/(-1/2)

f(1/2) =
5/2

f(-1) = (1 - 3 - 3)/(-1 -
1)

f(-1) = -5/-2

f(-1) =
5/2

The sum of the local extremes of the function
is:

f(1/2) + f(-1) = 5/2 +
5/2

f(1/2) + f(-1) =
5