Absolute min and max

Question

f(x)= 2sqrt(3+x^2) -x [ 0,2]

Answer 1

f(x) = 2√(3+x²) - x

 

f(x) is continuous on [0,2]

 

So, f achieves an absolute max and absolute min on the interval.  An absolute extremum occurs at either a critical point in (0,2) or at an endpoint.  

 

f'(x) = (3+x²)^-1/2(2x) - 1 = 2x/√(3+x²) - 1

 

                                   = [2x - √(3+x²)]/√(3+x²) = 0

 

              2x = √(3+x²)

 

              4x² = 3+x²

 

              3x² = 3      x² = 1    x = 1 or -1

 

                                              (discard -1 since it's not in the                                                                interval)

 

f(1) = 2√4 - 1 = 3 ← absolute minimum value 

 

f(0) = 2√3 ≈ 3.464 ← absolute maximum value is 2√3 

 

f(2) = 2√7 - 2 ≈ 3.29