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# y=x+1/x Find the minimum value of y if x>0.

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y=x+1/x Find the minimum value of y if x>0.

Guest Apr 28, 2015

#1
+78574
+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
Sort:

#1
+78574
+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

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