My question is similar to $xy=1 \implies $minimum $x+y=$?. Given that xy = 1 and x, y are real numbers we need to find minimum of x + y . So by doing the first derivative, I find that minimum comes at x = y = 1; that is minimum is 2. But take x = -100 and y = -1/100. Now xy = 1 but x + y = -100.01 ; which is much less than 2. Similarly we can take a lot of negative numbers and prove that 2 is never the minimum, but why did the derivative method gave me 2 as minimum? What was the mistake?
Asked
Active
Viewed 37 times
0
-
1Because it is a local minimum. – lulu Mar 12 '20 at 11:18
-
So 2 is a local minimum? – J Arun Mani Mar 12 '20 at 11:20
-
2Yes. Setting the derivative to $0$ just gives you local critical points, not absolute max and min. – lulu Mar 12 '20 at 11:21
-
A silly mistake by me... – J Arun Mani Mar 12 '20 at 11:28
-
1Yeah I used the second derivative test to approve that x = 1 is the minimum (now learnt that it is a local minima) – J Arun Mani Mar 12 '20 at 11:31
1 Answers
1
Well, when we find stationary points, we find local maxima and local minima and not the absolute maxima or absolute minima. For the function $f(x)=x+\frac{1}{x}$ you have a locan minima at $x=1$ which is higher than the local maxima at $x=-1$
So as seen in the graph, in its neighbourhood it is minima at $x=1$ and maxima at $x=-1$, but they are not global maxima and minima
user577215664
- 40,625
Mathsmerizing
- 2,003