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Question
Find the interval in which the following function are increasing or decreasing \[f\left( x \right) = \left\{ x(x - 2) \right\}^2\] ?
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Solution
\[\text { When } \left( x - a \right)\left( x - b \right)>0 \text { with }a < b, x < a \text { or }x>b.\]
\[\text { When } \left( x - a \right)\left( x - b \right)<0 \text { with } a < b, a < x < b .\]
\[f\left( x \right) = \left\{ x\left( x - 2 \right) \right\}^2 \]
\[ = \left( x^2 - 2x \right)^2 \]
\[ = x^4 + 4 x^2 - 4 x^3 \]
\[f'\left( x \right) = 4 x^3 + 8x - 12 x^2 \]
\[ = 4x \left( x^2 - 3x + 2 \right)\]
\[ = 4x \left( x - 1 \right)\left( x - 2 \right)\]
\[\text { Here, 0, 1 and 2 are the critical points}.\]
\[\text { The possible intervals are }\left( - \infty , 0 \right),\left( 0, 1 \right),\left( 1, 2 \right)\text { and }\left( 2, \infty \right).\]
\[\text { For f(x) to be increasing, we must have }\]
\[f'\left( x \right) > 0\]
\[ \Rightarrow 4x \left( x - 1 \right)\left( x - 2 \right) > 0\]
\[ \Rightarrow \left( x - 1 \right)\left( x - 2 \right) > 0\]
\[ \Rightarrow x \in \left( 0, 1 \right) \cup \left( 2, \infty \right) \]
\[\text { So,f(x)is increasing on x } \in \left( 0, 1 \right) \cup \left( 2, \infty \right) . \]
\[\text { For } f(x)\text { to be decreasing, we must have } \]
\[f'(x) < 0\]
\[ \Rightarrow 4x\left( x - 1 \right)\left( x - 2 \right) < 0\]
\[ \Rightarrow x\left( x - 1 \right)\left( x - 2 \right) < 0\]
\[ \Rightarrow x \in \left( - \infty , 0 \right) \cup \left( 1, 2 \right)\]
\[\text { So, f(x) is decreasing on x } \in \left( - \infty , 0 \right) \cup \left( 1, 2 \right) .\]
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