PUC Karnataka Science Class 12Department of Pre-University Education, Karnataka
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# Solution for F(X) = X3 − 6x2 + 9x + 15 - PUC Karnataka Science Class 12 - Mathematics

#### Question

f(x) = x3$-$ 6x2 + 9x + 15

#### Solution

$\text { Given }: f\left( x \right) = x^3 - 6 x^2 + 9x + 15$

$\Rightarrow f'\left( x \right) = 3 x^2 - 12x + 9$

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

$f'\left( x \right) = 0$

$\Rightarrow 3 x^2 - 12x + 9 = 0$

$\Rightarrow x^2 - 4x + 3 = 0$

$\Rightarrow \left( x - 1 \right)\left( x - 3 \right) = 0$

$\Rightarrow x = 1 \text { and } 3$

$\text { Thus, x = 1 and x = 3 are the possible points of local maxima or local minima } .$

$\text { Now,}$

$f''\left( x \right) = 6x - 12$

$\text { At }x = 1:$

$f''\left( 1 \right) = 6\left( 1 \right) - 12 = - 6 < 0$

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

$\text { The local maximum value is given by }$

$f\left( 1 \right) = 1^3 - 6 \left( 1 \right)^2 + 9 \times 1 + 15 = 19$

$\text { At }x = 3:$

$f''\left( 3 \right) = 6\left( 3 \right) - 12 = 6 > 0$

$\text { So, x = 3 is the point of local minimum }.$

$\text { The local minimum value is given by }$

$f\left( 3 \right) = 3^3 - 6 \left( 3 \right)^2 + 9 \times 3 + 15 = 15$

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Solution F(X) = X3 − 6x2 + 9x + 15 Concept: Graph of Maxima and Minima.
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