Sachin9 wrote:
So , if we have a inequality, change it with '=' , find the transition point and test using values that are on either side of transition point ..
But what analysis can we make out of the change by pluggin in the nos frm either side of the transition point?
You get the relation between the two expressions in various ranges of values.
To give an example:
To compare two terms e.g. \(x^2\) and \(2x\), we should focus on the points where they are equal. \(x^2 = 2x\) holds when \(x = 2\).
When \(x < 2, x^2 < 2x\)
When \(x > 2, x^2 > 2x\)
We have two ranges here x < 2 and x > 2. We know the relation between the two expressions in the two ranges.
Similarly \(1/x = x^2\) when \(x = 1\)
When \(x < 1, 1/x > x^2\).
When \(x > 1, 1/x > x^2\)
Going on, \(1/x = 2x\) when \(x = 1/\sqrt{2}\)
When \(x < 1/\sqrt{2}, 1/x > 2x\)
When \(x > 1/\sqrt{2}, 1/x < 2x\)
So now you have three ranges in which the relation between the three expressions will be different:
\(x < 1/\sqrt{2}\)
\(1/\sqrt{2} < x < 1\)
\(x > 1\)
If \(x < 1/\sqrt{2}\),
\(1/x > 2x, 1/x > x^2\) and \(x^2 < 2x\)
So \(x^2 < 2x < 1/x\) is possible.
If \(1/\sqrt{2} < x < 1\)
\(1/x < 2x, 1/x > x^2\)
So \(x^2 < 1/x < 2x\) is possible.
If \(x > 1\)
\(1/x < 2x, 1/x > x^2\)
So \(x^2 < 1/x < 2x\) is possible. (Same as above)
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