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Question
Find the interval in which the following function are increasing or decreasing \[f\left( x \right) = \frac{x^4}{4} + \frac{2}{3} x^3 - \frac{5}{2} x^2 - 6x + 7\] ?
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Solution
\[\text { When } \left( x - a \right)\left( x - b \right)>0 \text { with }a < b, x < a \text { or }x>b.\]
\[\text { When } \left( x - a \right)\left( x - b \right)<0 \text { with } a < b, a < x < b .\]
\[f\left( x \right) = \frac{x^4}{4} + \frac{2}{3} x^3 - \frac{5}{2} x^2 - 6x + 7\]
\[ = \frac{3 x^4 + 8 x^3 - 30 x^2 - 72x + 84}{12}\]
\[f'\left( x \right) = \frac{12 x^3 + 24 x^2 - 60x - 72}{12}\]
\[ = \left( x^3 + 2 x^2 - 5x - 6 \right)\]
\[ = \left( x + 1 \right)\left( x^2 + x - 6 \right)\]
\[ = \left( x + 1 \right)\left( x - 2 \right)\left( x + 3 \right)\]
\[\text { Here }, -1, 2 \text { and } -3 \text { are the critical points }.\]
\[\text { The possible intervals are }\left( - \infty - 3 \right),\left( - 3, - 1 \right),\left( - 1, 2 \right)and\left( 2, \infty \right).\]
\[\text { For }f(x) \text { to be increasing, we must have }\]
\[f'\left( x \right) > 0\]
\[ \Rightarrow \left( x + 1 \right)\left( x - 2 \right)\left( x + 3 \right) > 0\]
\[ \Rightarrow x \in \left( - 3, - 1 \right) \cup \left( 2, \infty \right)\]
\[\text { So },f(x)\text { is increasing on } x \in \left( - 3, - 1 \right) \cup \left( 2, \infty \right) . \]
\[\text { For }f(x) \text { to be decreasing, we must have }\]
\[f'\left( x \right) < 0\]
\[ \Rightarrow \left( x + 1 \right)\left( x - 2 \right)\left( x + 3 \right) < 0\]
\[ \Rightarrow x \in \left( - \infty - 3 \right) \cup \left( - 1, 2 \right) \left[ \text { From eq }. (1) \right]\]
\[\text { So,}f(x)\text { is decreasing on x } \in \left( - \infty - 3 \right) \cup \left( - 1, 2 \right) .\]
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