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Question
Find the interval in which the following function are increasing or decreasing \[f\left( x \right) = 3 x^4 - 4 x^3 - 12 x^2 + 5\] ?
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Solution
\[\text { When } \left( x - a \right)\left( x - b \right)>0 \text { with }a < b, x < a \text { or }x>b.\]
\[\text { When } \left( x - a \right)\left( x - b \right)<0 \text { with } a < b, a < x < b .\]
\[f\left( x \right) = 3 x^4 - 4 x^3 - 12 x^2 + 5\]
\[f'\left( x \right) = 12 x^3 - 12 x^2 - 24x\]
\[ = 12x\left( x^2 - x - 2 \right)\]
\[ = 12x\left( x + 1 \right)\left( x - 2 \right)\]
\[\text { Here }, x = 0,x = - 1\text { and x = 2 are the critical points }.\]
\[\text { The possible intervals are }\left( - \infty , - 1 \right),\left( - 1, 0 \right),\left( 0, 2 \right)\text { and }\left( 2, \infty \right). .....(1)\]
\[\text { For f(x) to be increasing, we must have }\]
\[f'\left( x \right) > 0\]
\[ \Rightarrow 12x\left( x + 1 \right)\left( x - 2 \right) > 0 \left[ \text { Since }, 12 > 0, 12x\left( x + 1 \right)\left( x - 2 \right) > 0 \Rightarrow x\left( x + 1 \right)\left( x - 2 \right) > 0 \right]\]
\[ \Rightarrow x\left( x + 1 \right)\left( x - 2 \right) > 0\]
\[ \Rightarrow x \in \left( - 1, 0 \right) \cup \left( 2, \infty \right) \left[ \text { From eq. } (1) \right]\]
\[\text { So,f(x)is increasing on x } \in \left( - 1, 0 \right) \cup \left( 2, \infty \right) .\]
\[\text { For f(x) to be decreasing, we must have }\]
\[f'\left( x \right) < 0\]
\[ \Rightarrow 12x\left( x + 1 \right)\left( x - 2 \right) < 0 \left[ \text { Since }12 > 0, 12x\left( x + 1 \right)\left( x - 2 \right) < 0 \Rightarrow x\left( x + 1 \right)\left( x - 2 \right) < 0 \right]\]
\[ \Rightarrow x \left( x + 1 \right)\left( x - 2 \right) < 0\]
\[ \Rightarrow x \in \left( - \infty , - 1 \right) \cup \left( 0, 2 \right) \left[ \text { From eq. } (1) \right]\]
\[\text { So,f(x)is decreasing on x }\in \left( - \infty , - 1 \right) \cup \left( 0, 2 \right) .\]
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