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?
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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