Approximate the coordinates of the relative minimum point of the graph of f

Question

Let f(x) = (3x² + 2x + 1)/x for x > 0

Approximate the coordinates of the relative minimum point of the graph of f.  Solve with a graphing calculator.

Answer 1

Simplify f(x).

 

f(x) = 3x + 2 + x^-1

 

 

Set the derivative of this equal to zero.

 

3 - x^-2 =  0

 

(3x² - 1) / x² = 0

 

 

Set the numerator equal to zero.

 

3x² - 1 = 0

 

3x² = 1

 

  x = ±1/√3

 

Since x is positive, the only critical value accepted is

 

x = 0.57

 

Evaluate the derivative around this critical value.  Since x² is always positive, we only care about the sign of the numerator.

 

f'(1/2) = negative

f'(1) = positive

 

x=1/√3 contains a relative minimum point.

Answer 2

y = 3x + 2 + x^-1 after doing the division, and therefore y' = 3 + (-1)x^-2

to find critical points, y' = 0 must be solved.

 

3 = x^-2

1/3 = x²

x = 1 / √3 (since x>0 only)

 

Etc...

 

This is where a minimum is, because y" = 2x^-3 which is > 0 for x > 0 and thus concavity is upward at this critical point, meaning it's a local (relative) minimum.