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

*print*Print*list*Cite

### 2 Answers

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.**

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**