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Question
f(x) = x4 \[-\] 62x2 + 120x + 9.
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Solution
\[\text { Given: }f\left( x \right) = x^4 - 62 x^2 + 120x + 9\]
\[ \Rightarrow f'\left( x \right) = 4 x^3 - 124x + 120\]
\[\text {For the local maxima or minima, we must have }\]
\[ f'\left( x \right) = 0\]
\[ \Rightarrow 4 x^3 - 124x + 120 = 0\]
\[ \Rightarrow x^3 - 31x + 30 = 0\]
\[ \Rightarrow \left( x - 1 \right)\left( x^2 + x - 30 \right) = 0\]
\[ \Rightarrow \left( x - 1 \right)\left( x + 6 \right)\left( x - 5 \right) = 0\]
\[ \Rightarrow x = 1, 5 \text { and } - 6\]
\[\text { Thus, x = 1, x = 5 and x = - 6 are the possible points of local maxima or local minima} . \]
\[\text { Now,} \]
\[f''\left( x \right) = 12 x^2 - 124\]
`"At "x = 1 :-`
\[f''\left( 1 \right) = 12 \left( 1 \right)^2 - 124 = - 112 < 0\]
\[\text { So, x = 1 is the point of local maximum} . \]
\[\text { The local maximum value is given by }\]
\[f\left( 1 \right) = 1^4 - 62 \left( 1 \right)^2 + 120 \times 1 + 9 = 68\]
\[\text { At } x = 5: \]
\[ f''\left( 5 \right) = 12 \left( 5 \right)^2 - 124 = 176 > 0\]
\[\text { So, x = 5 is the point of local minimum }. \]
\[\text { The local minimum value is given by }\]
\[f\left( 5 \right) = 5^4 - 62 \left( 5 \right)^2 + 120 \times 5 + 9 = - 316\]
\[\text { At }x = - 6: \]
\[ f''\left( - 6 \right) = 12 \left( - 6 \right)^2 - 124 = 308 > 0\]
\[\text { So, x = - 6 is the point of local maximum }. \]
\[\text { The local minimum value is given by } \]
\[f\left( - 6 \right) = \left( - 6 \right)^4 - 62 \left( - 6 \right)^2 + 120 \times \left( - 6 \right) + 9 = - 1647\]
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