Absolute min and max

Question

f(x)= 2(3+x^2)^(1/2) -x on the interval [0,2]

Answer 1

Set the derivative of f(x) equal to zero.

 

[2(1/2)(3 + x²)^-1/2(2x)] - 1 = 0

 

 

Solve for x.

 

[2x(3 + x²)^-1/2] - 1 = 0

 

[2x / √(3 + x²)] - 1 = 0

 

2x / √(3 + x²) = 1

 

2x = √(3 + x²) 

 

4x² = 3 + x²

 

3x² - 3 = 0

 

3(x² - 1) = 0

 

x = -1         and          x = 1

 

 

We reject  x=-1 because it does not make      2x=√(3+x²)         true.

  

So your local critical point is located as x=1.

 

 

Now just evaluate f(0), f(1), and f(2).  Compare these values.  The lowest of the values is the absolute minimum.  The highest of these values is the absolute maximum.

Answer 2

Differentiate and set f' = 0 in order to find any critical point.

 

To find abs min/max, also evaluate f(0) and f(2), which are endpoints.

 

Then pick the smallest and the largest of f(c.p.), f(0) and f(2).

 

f' = 2(½)(2x)(3+x²)^-½ - 1