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Thursday, October 29, 2015

UTM Ideals Varieties and Algorithm - Chapter 1 Section 2 Exercise 2

Problem:


Solution:

Note that $ y^2 $ is non-negative, therefore we only need to focus on range of $ x $ such that $ x(x - 1)(x - 2) $ is non-negative.

The only two such ranges is $ [0, 1] $ and $ [2, +\infty) $.

Next, we look at the range $ [0, 1] $, the curve start from 0 and go back to 0, therefore it must has a turning point, taking the first derivative we get

$ \begin{eqnarray*} f(x) &=& x(x - 1)(x - 2) \\ f'(x) &=& (x - 1)(x - 2) + x(x - 1) + x(x - 2) \\ &=& (x^2 - 3x + 2) + (x^2 - x) + (x^2 - 2x) \\ &=& 3x^2 - 6x + 2 \\ &=& 3(x^2 - 2x + 1) - 1 \end{eqnarray*} $

Therefore the curve turn at $ x = -\sqrt{\frac{1}{3}} + 1 = 0.42... $.

Last but not least, the curve is growing cubic fast in $ [2, +\infty) $, it is easy to sketch there.




It looks like a fish, isn't it?

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