Question

f(x) = x⁴ \[-\] 62x² + 120x + 9.

Answer 1

\[\text { Given: }f\left( x \right) = x^4 - 62 x^2 + 120x + 9\]

\[ \Rightarrow f'\left( x \right) = 4 x^3 - 124x + 120\]

\[\text {For the local maxima or minima, we must have }\]

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

\[ \Rightarrow 4 x^3 - 124x + 120 = 0\]

\[ \Rightarrow x^3 - 31x + 30 = 0\]

\[ \Rightarrow \left( x - 1 \right)\left( x^2 + x - 30 \right) = 0\]

\[ \Rightarrow \left( x - 1 \right)\left( x + 6 \right)\left( x - 5 \right) = 0\]

\[ \Rightarrow x = 1, 5 \text { and } - 6\]

\[\text { Thus, x = 1, x = 5 and x = - 6 are the possible points of local maxima or local minima} . \]

\[\text { Now,} \]

\[f''\left( x \right) = 12 x^2 - 124\]

`"At "x = 1 :-`

\[f''\left( 1 \right) = 12 \left( 1 \right)^2 - 124 = - 112 < 0\]

\[\text { So, x = 1 is the point of local maximum} . \]

\[\text { The local maximum value is given by }\]

\[f\left( 1 \right) = 1^4 - 62 \left( 1 \right)^2 + 120 \times 1 + 9 = 68\]

\[\text { At } x = 5: \]

\[ f''\left( 5 \right) = 12 \left( 5 \right)^2 - 124 = 176 > 0\]

\[\text { So, x = 5 is the point of local minimum }. \]

\[\text { The local minimum value is given by }\]

\[f\left( 5 \right) = 5^4 - 62 \left( 5 \right)^2 + 120 \times 5 + 9 = - 316\]

\[\text { At }x = - 6: \]

\[ f''\left( - 6 \right) = 12 \left( - 6 \right)^2 - 124 = 308 > 0\]

\[\text { So, x = - 6 is the point of local maximum }. \]

\[\text { The local minimum value is given by } \]

\[f\left( - 6 \right) = \left( - 6 \right)^4 - 62 \left( - 6 \right)^2 + 120 \times \left( - 6 \right) + 9 = - 1647\]