15) Let I be any interval disjoint from [–1, 1]. Prove that the function f given by f ( x) = x + 1/x  is increasing on I.

Answers (1)

Given function is,
f ( x) = x + 1/x
f^{'}(x) = 1 - \frac{1}{x^2}
Now,  
f^{'}(x) = 0\\ 1 - \frac{1}{x^2} = 0\\ x^{2} = 1\\ x = \pm1

So, intervals are from  (-\infty,-1), (-1,1) \ and \ (1,\infty)
In interval (-\infty,-1), (1,\infty)  ,  \frac{1}{x^2} < 1 \Rightarrow 1 - \frac{1}{x^2} > 0
f^{'}(x) > 0
Hence, f ( x) = x + 1/x is increasing   in  interval  (-\infty,-1)\cup (1,\infty)
In interval (-1,1) ,  \frac{1}{x^2} > 1 \Rightarrow 1 - \frac{1}{x^2} < 0
f^{'}(x) < 0
Hence, f ( x) = x + 1/x is decreasing   in  interval  (-1,1)
Hence, the function f given by f ( x) = x + 1/x  is increasing on I  disjoint from [–1, 1]

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