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Question
Find the interval in which the following function are increasing or decreasing f(x) = 6 + 12x + 3x2 − 2x3 ?
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Solution
\[\text { When } \left( x - a \right)\left( x - b \right)>0 \text { with }a < b, x < a \text { or }x>b.\]
\[\text { When } \left( x - a \right)\left( x - b \right)<0 \text { with } a < b, a < x < b .\]
\[f\left( x \right) = 6 + 12x + 3 x^2 - 2 x^3 \]
\[f'\left( x \right) = 12 + 6x - 6 x^2 \]
\[ = - 6 \left( x^2 - x - 2 \right)\]
\[ = - 6 \left( x - 2 \right)\left( x + 1 \right)\]
\[\text { For }f(x) \text { to be increasing, we must have }\]
\[f'\left( x \right) > 0\]
\[ \Rightarrow - 6 \left( x - 2 \right)\left( x + 1 \right) > 0\]
\[ \Rightarrow \left( x - 2 \right)\left( x + 1 \right) < 0 \left[ \text { Since } - 6 < 0, - 6 \left( x - 2 \right)\left( x + 1 \right) > 0 \Rightarrow \left( x - 2 \right)\left( x + 1 \right) < 0 \right]\]
\[ \Rightarrow - 1 < x < 2 \]
\[ \Rightarrow x \in \left( - 1, 2 \right)\]
\[\text { So },f(x)\text { is increasing on} \left( - 1, 2 \right) . \]
\[\text { For }f(x) \text { to be decreasing, we must have }\]
\[f'\left( x \right) < 0\]
\[ \Rightarrow - 6 \left( x - 2 \right)\left( x + 1 \right) < 0\]
\[ \Rightarrow \left( x - 2 \right)\left( x + 1 \right) > 0 \left[ \text { Since } - 6 < 0, - 6 \left( x - 2 \right)\left( x + 1 \right) < 0 \Rightarrow \left( x - 2 \right)\left( x + 1 \right) > 0 \right]\]
\[ \Rightarrow x <\text{ - 1 or x} > 2 \]
\[ \Rightarrow x \in \left( - \infty , - 1 \right) \cup \left( 2, \infty \right)\]
\[\text { So },f(x)\text { is decreasing on }\left( - \infty , - 1 \right) \cup \left( 2, \infty \right) .\]
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