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# Solution for Find the Interval in Which the Following Function Are Increasing Or Decreasing F ( X ) = { X ( X − 2 ) } 2 ? - CBSE (Commerce) Class 12 - Mathematics

ConceptIncreasing and Decreasing Functions

#### Question

Find the interval in which the following function are increasing or decreasing $f\left( x \right) = \left\{ x(x - 2) \right\}^2$ ?

#### Solution

$\text { When } \left( x - a \right)\left( x - b \right)>0 \text { with }a < b, x < a \text { or }x>b.$

$\text { When } \left( x - a \right)\left( x - b \right)<0 \text { with } a < b, a < x < b .$

$f\left( x \right) = \left\{ x\left( x - 2 \right) \right\}^2$

$= \left( x^2 - 2x \right)^2$

$= x^4 + 4 x^2 - 4 x^3$

$f'\left( x \right) = 4 x^3 + 8x - 12 x^2$

$= 4x \left( x^2 - 3x + 2 \right)$

$= 4x \left( x - 1 \right)\left( x - 2 \right)$

$\text { Here, 0, 1 and 2 are the critical points}.$

$\text { The possible intervals are }\left( - \infty , 0 \right),\left( 0, 1 \right),\left( 1, 2 \right)\text { and }\left( 2, \infty \right).$

$\text { Forf(x) to be increasing, we must have }$

$f'\left( x \right) > 0$

$\Rightarrow 4x \left( x - 1 \right)\left( x - 2 \right) > 0$

$\Rightarrow \left( x - 1 \right)\left( x - 2 \right) > 0$

$\Rightarrow x \in \left( 0, 1 \right) \cup \left( 2, \infty \right)$

$\text { So,f(x)is increasing on x } \in \left( 0, 1 \right) \cup \left( 2, \infty \right) .$

$\text { For } f(x)\text { to be decreasing, we must have }$

$f'(x) < 0$

$\Rightarrow 4x\left( x - 1 \right)\left( x - 2 \right) < 0$

$\Rightarrow x\left( x - 1 \right)\left( x - 2 \right) < 0$

$\Rightarrow x \in \left( - \infty , 0 \right) \cup \left( 1, 2 \right)$

$\text { So, f(x) is decreasing on x } \in \left( - \infty , 0 \right) \cup \left( 1, 2 \right) .$

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Solution Find the Interval in Which the Following Function Are Increasing Or Decreasing F ( X ) = { X ( X − 2 ) } 2 ? Concept: Increasing and Decreasing Functions.
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