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

Expert Answers
justaguide eNotes educator| Certified Educator

To maximize the value of f(x) = (x-3)/ (2x^2 - 5), we find the first derivative and solve for x given by f'(x) = 0

f(x) = (x - 3)*(2x^2 - 5)^-1

f'(x) = -(x - 3)*4x*(2x^2 - 5)^-2 + (2x^2 - 5)^-1

-(x - 3)*4x*(2x^2 - 5)^-2 + (2x^2 - 5)^-1 = 0

=> (x - 3)*4x = (2x^2 - 5)

=> 4x^2 - 12x = 2x^2 - 5

=> 4x^2 - 14x + 5 = 0

x1 = 14/8 + sqrt (196 - 80)/8

=> 7/4 + sqrt (116)/8

x2 = 7/4 - sqrt (116)/8

The maximum value is at the point where x = 7/4 - sqrt (116)/8 and is equal to 0.5554

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)