What is the maximum value of f(x) = (x-3)/ (2x^2 - 5) 

Expert Answers
hala718 eNotes educator| Certified Educator

f(x) = (x-3)/ (2x^2 - 5)

First we will differentiate:

f(x) = u/v such that:

u = (x-3)  ==>   u' = 1

v = (2x^2 - 5) ==> v' = 4x

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

               = ( (2x^2 - 5) - ( 4x(x-3) / ( ax^2 - 5)^2

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

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

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

   Now we will find the zeros:

==> ( 2x^2 - 12 x + 5) = 0

==> x1= ( 12 + sqrt( 144- 4*2*5) / 4

             = 12 + sqrt( 104) / 4

              = 3 +  sqrt26 / 2

==> x2= 3 - sqrt26/ 2

giorgiana1976 | Student

To calculate the maximum value of the given function, first we'll have to determine the critical points.

For this reason, we'll determine the first derivative of the function, knowing that the critical point are the roots of the first derivative.

Since the function is a ratio, we'll apply the quotient rule:

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

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

We'll remove the brackets:

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

We'll combine like terms:

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

We'll put f'(x) = 0

Since the denominator is always positive, for any value of x, we'll find the roots of the numerator:

2x^2 - 12x + 5 = 0

We'll apply the quadratic formula:

x1 = [12+sqrt(144-40)]/4

The critical values of the function are:

x1 = (6+sqrt26)/2

x2 = (6-sqrt26)/2

The extreme values of the function are:

f(x1) = ((6+sqrt26)/23)/(2((6+sqrt26)/2)^2 - 5)

f(x2) = ((6-sqrt26)/23)/(2((6-sqrt26)/2)^2 - 5)