Consider the inequality \(\frac{1}{x} < 2\). It says to consider 2 cases.
You already found that \(x > \frac{1}{2}\). How do you find the other case?
Answer (2)
[\frac{1}{x}<2|\cdot x^2\ x<2x^2\ 2x^2-x>0\ x(2x-1)>0\ x\in(-\infty,0)\cup(\frac{1}{2},\infty)
]
To find the other case for the inequality x 1 < 2 , we consider negative values of x . This leads to the conclusion that valid solutions are x < 0 or \frac{1}{2}"> x > 2 1 .
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