Problem:
Solution:
Note that y2 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,+∞).
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
f(x)=x(x−1)(x−2)f′(x)=(x−1)(x−2)+x(x−1)+x(x−2)=(x2−3x+2)+(x2−x)+(x2−2x)=3x2−6x+2=3(x2−2x+1)−1
Therefore the curve turn at x=−√13+1=0.42....
Last but not least, the curve is growing cubic fast in [2,+∞), it is easy to sketch there.
It looks like a fish, isn't it?
Solution:
Note that y2 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,+∞).
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
f(x)=x(x−1)(x−2)f′(x)=(x−1)(x−2)+x(x−1)+x(x−2)=(x2−3x+2)+(x2−x)+(x2−2x)=3x2−6x+2=3(x2−2x+1)−1
Therefore the curve turn at x=−√13+1=0.42....
Last but not least, the curve is growing cubic fast in [2,+∞), it is easy to sketch there.
It looks like a fish, isn't it?
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