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Question
Find the value(s) of x for which y = [x(x − 2)]2 is an increasing function.
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Solution
f(x)=[x(x−2)]2
f'(x)=2[x(x−2)]{x−2+x}
f'(x)=4x(x−2)(x−1)
At critical point, f'(x)=0
4x(x−2)(x−1)=0
⇒x=0,1,2
| Interval | f'(x)=4x(x−1)(x−2) |
Result |
| (−∞,0) | f'(−1)=4(−1)(−2)(−3)=−24<0 | Decreasing |
| (0,1) | f'(1/2)=4(1/2)(−1/2)(−3/2)=3/2>0 | Increasing |
| (1,2) | f'(3/2)=4(3/2)(1/2)(−1/2)=−3/2<0 | Decreasing |
| (2,∞) | f'(3)=4(3)(2)(1)=24>0 | Increasing |
So, the function is increasing in the interval (0,1)∪(2,∞).
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