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Question
Find the interval in which the following function are increasing or decreasing \[f\left( x \right) = \frac{3}{10} x^4 - \frac{4}{5} x^3 - 3 x^2 + \frac{36}{5}x + 11\] ?
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Solution
\[\text { When } \left( x - a \right)\left( x - b \right)>0 \text { with }a < b, x < a \text { or }x>b.\]
\[\text { When } \left( x - a \right)\left( x - b \right)<0 \text { with } a < b, a < x < b .\]
\[f\left( x \right) = \frac{3}{10} x^4 - \frac{4}{5} x^3 - 3 x^2 + \frac{36}{5}x + 11\]
\[ = \frac{3 x^4 - 8 x^3 - 30 x^2 + 72x + 110}{10}\]
\[f'\left( x \right) = \frac{12 x^3 - 24 x^2 - 60x + 72}{10}\]
\[ = \frac{12}{10}\left( x^3 - 2 x^2 - 5x + 6 \right)\]
\[ = \frac{\left( x - 1 \right)\left( x^2 - x - 6 \right)}{10}\]
\[ = \frac{12}{10}\left( x - 1 \right)\left( x + 2 \right)\left( x - 3 \right)\]
\[\text { Here }, 1, 2 \text { and } 3 \text { are the critical points } . \]
\[\text { The possible intervals are }\left( - \infty - 2 \right),\left( - 2, 1 \right),\left( 1, 3 \right)\text { and }\left( 3, \infty \right).\]
\[\text { For }f(x)\text { to be increasing, we must have }\]
\[f'\left( x \right) > 0\]
\[ \Rightarrow \frac{12}{10}\left( x - 1 \right)\left( x + 2 \right)\left( x - 3 \right) > 0\]
\[ \Rightarrow \left( x - 1 \right)\left( x + 2 \right)\left( x - 3 \right) > 0\]
\[ \Rightarrow x \in \left( - 2, 1 \right) \cup \left( 3, \infty \right)\]
\[\text { So },f(x)\text { is increasing on } x \in \left( - 2, 1 \right) \cup \left( 3, \infty \right) . \]
\[\text { For }f(x)\text { to be decreasing, we must have }\]
\[f'\left( x \right) < 0\]
\[ \Rightarrow \frac{12}{10}\left( x - 1 \right)\left( x + 2 \right)\left( x - 3 \right) < 0\]
\[ \Rightarrow \left( x - 1 \right)\left( x + 2 \right)\left( x - 3 \right) < 0\]
\[ \Rightarrow x \in \left( - \infty - 2 \right) \cup \left( 1, 3 \right) \]
\[\text { So,}f(x)\text { is decreasing on } x \in \left( - \infty - 2 \right) \cup \left( 1, 3 \right) .\]
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