For the function f(x) = (x^2-4x+5)/(x+3). find the extreme values.

To find the extreme values of a function, we'll need to
determine the critical alues. The critical values are the roots of the first
derivative.

We'll differentiate the function to determine it's first
derivative.

We'll use the quotient
rule:

f'(x) = [(x^2-4x+5)'*(x+3) -
(x^2-4x+5)*(x+3)']/(x+3)^2

f'(x) = [(2x - 4)*(x+3) -
(x^2-4x+5)]/(x+3)^2

f'(x) = (2x^2 + 6x - 4x - 12 - x^2 + 4x -
5)/(x+3)^2

f'(x) = (x^2 + 6x -
17)/(x+3)^2

We'll put f'(x) = 0

Since
the denominator is always positive, no matter the x value is, we'll determine what values cancel
the numerator.

x^2 + 6x - 17 = 0

We'll
apply quadratic formula:

x1 = [-6+sqrt(36 +
68)]/2

x1 = (-6+sqrt104)/2

x1 =
(-6+2sqrt26)/2

x1 = -3+sqrt26

x2 =
-3-sqrt26

Since we know the critical values, we'll determine the
extreme values:

f(-3+sqrt26) = f(2.09) = (4.36-8.36+5)/(2.09+3) =
9/5.09 = 1.76

f(-3-sqrt26) = f (-8.09) = (65.44+32.36+5)/(-5.09) =
102.8/-5.09 = -20.19

The extreme value are represented
by the pairs: (-3+sqrt26 ; 1.76) and (-3-sqrt26 ; -20.19).