#### Provide solution for RD Sharma maths class 12 chapter Maxima and Minima exercise 17.4 question 2

Maximum value $=89 \text { at } x=3 \text { in }[1,3]$

Maximum value $=139 \text { at } x=2 \text { in }[-3,-1]$

Hint:

check the value of f(x) at end points where f ‘(x) = 0

Given:

$f(x)=2 x^{3}-24 x+107 \text { in }[1,3]$     ,   $4 \text { in }[-3,-1]$

Explanation:

$f^{\prime}(x)=6 x^{2}-24$

\begin{aligned} &f^{\prime}(x)=0 \\ &6 x^{2}-24=0 \\ &x^{2}-4=0 \end{aligned}

\begin{aligned} &(x-2)(x+2)=0 \quad\quad\quad\quad\left[a^{2}-b^{2}=(a-b)(a+b)\right] \\ &x=2, x=-2 \end{aligned}

Now,  we check first in the internal (1,3)

\begin{aligned} &f(1)=2 \times 1-24 \times 1+107=109-24=85 \\ &f(2)=2 \times\left(2^{3}\right)-24 \times 2+107=16-42+107=75 \\ &f(3)=2 \times\left(3^{3}\right)-24 \times 3+107=54-72+107=89 \end{aligned}

Hence,

Absolute Maximum $=89 \text { at } x=3 \text { in }[1,3]$

Now, we check in the internal $[-3,-1]$

\begin{aligned} &f(-3)=2 \times(-3)^{3}-24 \times(-3)+107 \\\\ &=-54+72+107=125 \\\\ &f(-2)=2 \times(-2)^{3}-24 \times(-2)+107 \end{aligned}

\begin{aligned} &=-16+42+107=139 \\\\ &f(-1)=2 \times(-1)^{3}-24 \times(-1)+107 \\\\ &=-2+24+107=129 \end{aligned}

Absolute Maximum $=139 \text { at } x=-2$