Find the value(s) of x for which y = [x(x - 2)]^2 is an increasin… - 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 $y = [x(x - 2)]^2= [x^2 - 2x]^2.$
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) $
$= 2(x^2 - 2x) (2x - 2)$
$= 4x(x - 2) (x - 1)$
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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