Find the maximum and minimum values of the function f(x)=x-((8x)/(x+2) on the interval [0,3].

Question

Find the maximum and minimum values of the function f(x)=x−((8x)/(x+2) on the interval [0,3].

The minimum value =

The maximum value =

Answer 1

f(x) = x - 8x/(x+2) is continuous and differentiable on the interval [0,3], so the max and min occur at a critical point or at an endpoint.  

 

To find the critical numbers, set f'(x) = 0.

 

f'(x) = 1 - [8(x+2) - 8x]/(x+2)²

 

          = [(x+2)² -16]/(x+2)²

 

          = (x²+4x-12)/(x+2)² = [(x+6)(x-2)]/(x+2)²

 

f'(x) = 0 when x = -6 or x = 2

 

So, the only critical number in the interval [0,3] is 2.

 

Evaluate f(x) at x = 2, x = 0, and x = 3.

 

f(2) = -2

f(0) =  0

f(3) = -1.8

 

Absolute max = f(0) = 0

Absolute min = f(2) = -2