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Question
Find the maximum value of 2x3\[-\] 24x + 107 in the interval [1,3]. Find the maximum value of the same function in [ \[-\] 3, \[-\] 1].
Solution
\[\text { Given:} f\left( x \right) = 2 x^3 - 24x + 107\]
\[ \Rightarrow f'\left( x \right) = 6 x^2 - 24\]
\[\text { For a local maximum or a local minimum, we must have }\]
\[f'\left( x \right) = 0\]
\[ \Rightarrow 6 x^2 - 24 = 0\]
\[ \Rightarrow 6 x^2 = 24\]
\[ \Rightarrow x^2 = 4\]
\[ \Rightarrow x = \pm 2\]
\[\text { Thus, the critical points of f in the interval } \left[ 1, 3 \right] \text { are 1, 2 and 3 } . \]
\[\text { Now,} \]
\[ f\left( 1 \right) = 2 \left( 1 \right)^3 - 24\left( 1 \right) + 107 = 85\]
\[f\left( 2 \right) = 2 \left( 2 \right)^3 - 24\left( 2 \right) + 107 = 75\]
\[f\left( 3 \right) = 2 \left( 3 \right)^3 - 24\left( 3 \right) + 107 = 89\]
\[\text { Hence, the absolute maximum value when x = 3 in the interval } \left[ 1, 3 \right] is 89 . \]
\[\text { Again, the critical points of f in the interval } \left[ - 3, - 1 \right] \text {are - 1, - 2 and } - 3 . \]
\[\text { So }, \]
\[f\left( - 3 \right) = 2 \left( - 3 \right)^3 - 24\left( - 3 \right) + 107 = 125\]
\[f\left( - 2 \right) = 2 \left( - 2 \right)^3 - 24\left( - 2 \right) + 107 = 139\]
\[f\left( - 1 \right) = 2 \left( - 1 \right)^3 - 24\left( - 1 \right) + 107 = 129\]
\[\text { Hence, the absolute maximum value when } x = - 2 \text { is } 139 .\]
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