Maximum and minimum value of X+1/X
Since you are preparing for CAT/CSAT, you already know the range of values of x+1/x
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Isn’t it ??
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Well if you don’t, you have to devote more time to your studies.
The range of f(x) = x + 1/x is (-infinity, -2] U [2,infinity)
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But many a times we get confused how to get this range of values.
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Well easiest way to get the maximum and minimum values of any function f(x) is to differentiate. But the most important point to keep in mind is that the derivative method works when the function is defined at all point in the domain and is continuous.
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Lets check this function f(x) = x + 1/x.
We can clearly see that f(x) doesn’t exist at x=0 and its domain is (-infinity, -0) U (0,infinity)
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So how to deal with this situation.
Lets get back to derivative method again. Since we need the function which is continuous and defined at all points of the domain. We will break this function in 2 parts and then we will try to find its values in the domain.
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Case 1: when domain of f(x) belongs to (0,infinity)
f(x) = x +1/x
f'(x) = 1-1/x^2
f'(x) = 0 –> 1-1/x^2 = 0 –> x^2 -1 = 0 –> X=+/- 1. But x belongs to (0,infinity). S0 x =1
Now f”x = 2x^3 > 0 , since x belongs to (0,infinity).
So at x = 1 has minimum value and maximum value lies at infinity
f(1) = 1+1= 2
Hence for x belongs to (0,infinity), f(x) belongs to [2, infinity).
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Case 2: when domain of f(x) belongs to (-infinity,0)
f(x) = x +1/x
f'(x) = 1-1/x^2
f'(x) = 0 –> 1-1/x^2 = 0 –> x^2 -1 = 0 –> X=+/- 1. But x belongs to (0,infinity). S0 x = – 1
Now f”x = 2x^3 < 0 , since x belongs to (-inifinity,0).
So at x = -1 has maximum value and minimum value lies at infinity
f(-1) = -1-1= -2
Hence for x belongs to (-infinity,0), f(x) belongs to (-infinity, -2].
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From case 1 and Case 2, Range of f(x) is (-infinity, -2] U [2,infinity).
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Now how does the graph of this function would look like ?
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