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

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hala718's profile pic

hala718 | High School Teacher | (Level 1) Educator Emeritus

Posted on

Given f(x) = (x^2-4x+5) /(x+3)

We need to find the extreme values.

First we will need to find the derivatives zeros.

==> f(x) = u/v such that

u= x^2-4x +5 ==> u= 2x -4

v= x+3 ==> v' = 1

==> f'(x) = [ u'v-uv']/v^2

              = [ (2x-4)(x+3) - (x^2-4x+5)]/(x+3)62

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

             = (x^2 +6x -17) /(x+3)^2

==> f'(x) = (x^2 +6x -17) /(x+3)^2 = 0

==> x^2 +6x -17 = 0

==> x1= (-6+sqrt(36+4*17) / 2 = ( -6 + 2sqrt26)/2 = -3+sqrt26= 2.1 ( approx.)

==> x2= -3 -sqrt26= 8.1 ( approx.)

==> Then the function has extreme values at f(2.1) and f(8.1)

giorgiana1976's profile pic

giorgiana1976 | College Teacher | (Level 3) Valedictorian

Posted on

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).

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