PUC Karnataka Science Class 12Department of Pre-University Education, Karnataka
Share

Books Shortlist

# Solution for Find the Maximum Value of 2x3 − 24x + 107 in the Interval [1,3]. Find the Maximum Value of the Same Function in [ − 3, − 1]. - PUC Karnataka Science Class 12 - Mathematics

#### 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].

#### Solution

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

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

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

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

$\Rightarrow 6 x^2 - 24 = 0$

$\Rightarrow 6 x^2 = 24$

$\Rightarrow x^2 = 4$

$\Rightarrow x = \pm 2$

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

$\text { Now,}$

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

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

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

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

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

$\text { So },$

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

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

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

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

Is there an error in this question or solution?

#### Video TutorialsVIEW ALL [1]

Solution Find the Maximum Value of 2x3 − 24x + 107 in the Interval [1,3]. Find the Maximum Value of the Same Function in [ − 3, − 1]. Concept: Graph of Maxima and Minima.
S