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# Solution for F(X) = X4 − 62x2 + 120x + 9. - CBSE (Commerce) Class 12 - Mathematics

#### Question

f(x) = x4 $-$ 62x2 + 120x + 9.

#### 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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Solution F(X) = X4 − 62x2 + 120x + 9. Concept: Graph of Maxima and Minima.
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