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].

Answer 1

\[\text { Given:} f\left( x \right) = 2 x^3 - 24x + 107\]

\[ \Rightarrow f'\left( x \right) = 6 x^2 - 24\]

\[\text { For a local maximum or a local minimum, we must have }\]

\[f'\left( x \right) = 0\]

\[ \Rightarrow 6 x^2 - 24 = 0\]

\[ \Rightarrow 6 x^2 = 24\]

\[ \Rightarrow x^2 = 4\]

\[ \Rightarrow x = \pm 2\]

\[\text { Thus, the critical points of f in the interval } \left[ 1, 3 \right] \text { are 1, 2 and 3 } . \]

\[\text { Now,} \]

\[ f\left( 1 \right) = 2 \left( 1 \right)^3 - 24\left( 1 \right) + 107 = 85\]

\[f\left( 2 \right) = 2 \left( 2 \right)^3 - 24\left( 2 \right) + 107 = 75\]

\[f\left( 3 \right) = 2 \left( 3 \right)^3 - 24\left( 3 \right) + 107 = 89\]

\[\text { Hence, the absolute maximum value when x = 3 in the interval } \left[ 1, 3 \right] is 89 . \]

\[\text { Again, the critical points of f in the interval } \left[ - 3, - 1 \right] \text {are - 1, - 2  and } - 3 . \]

\[\text { So }, \]

\[f\left( - 3 \right) = 2 \left( - 3 \right)^3 - 24\left( - 3 \right) + 107 = 125\]

\[f\left( - 2 \right) = 2 \left( - 2 \right)^3 - 24\left( - 2 \right) + 107 = 139\]

\[f\left( - 1 \right) = 2 \left( - 1 \right)^3 - 24\left( - 1 \right) + 107 = 129\]

\[\text { Hence, the absolute maximum value when } x = - 2 \text { is } 139 .\]