#1**+5 **

y=x+1/x

Taking the derivative, we have....

y ' = 1 - 1/x^2 set to 0

1 - 1/x^2 = 0

1/x^2 = 1 and we have two critical points...... x = -1 or x = 1

Since x > 0 , let's take the second derivative

y " = 2/x^3 and evaluating this at x =1, we have 2/1^3 = 2 .......so......we have an "minimum" at (1 , 2). Note that this is just a minimum on the interval of interest, i.e., (0, ∞). This function actually has no absolute minimum (or maximum) on its domain.

Here's the graph..... https://www.desmos.com/calculator/bv0ty1o4ay

CPhill
Apr 28, 2015

#1**+5 **

Best Answer

y=x+1/x

Taking the derivative, we have....

y ' = 1 - 1/x^2 set to 0

1 - 1/x^2 = 0

1/x^2 = 1 and we have two critical points...... x = -1 or x = 1

Since x > 0 , let's take the second derivative

y " = 2/x^3 and evaluating this at x =1, we have 2/1^3 = 2 .......so......we have an "minimum" at (1 , 2). Note that this is just a minimum on the interval of interest, i.e., (0, ∞). This function actually has no absolute minimum (or maximum) on its domain.

Here's the graph..... https://www.desmos.com/calculator/bv0ty1o4ay

CPhill
Apr 28, 2015