Find the value(s) of x for which y = [x(x - 2)]2 is an increasing… - Vidyadip

Question

Find the value(s) of x for which y = [x(x - 2)]² is an increasing function.

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Answer

Given function is y = [x(x - 2)]²= [x² - 2x]².
Therefore,on differentiating both sides w.r.t x, we get,
$\frac { d y } { d x } = 2 \left( x ^ { 2 } - 2 x \right) \frac { d } { d x } \left( x ^ { 2 } - 2 x \right) $
= 2(x² - 2x) (2x - 2)
= 4x(x - 2) (x - 1)
Therefore,on putting $\frac { d y } { d x }$= 0, we get,
4x(x - 2)(x - 1) = 0$\Rightarrow$x = 0, 1 and 2.
Now, we find interval in which f(x) is strictly increasing or strictly decreasing.

Interval	f'(x) = 12x(x + 1)(x - 2)	Sign of f'(x)
(-$\infty$,0)	(-)(-)(-)	-ve
(0, 1)	(+)(-)(-)	+ve
(1, 2)	(+)(-)(+)	-ve
(2,$\infty$)	(+)(+)(+)	+ve

Therefore, y is strictly increasing in ( 0, 1 ) and (2, $\infty$).
Also, y is a polynomial function, so it continuous at x = 0, 1 and 2.
Hence, y is increasing in [ 0, 1] $ \cup $ [2, $\infty$).

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