(0, 9) ব্যবধিতে f(x)=x³-18x²+96x ফাংশনের  সর্বোচ্চ মান কত ?  

$\begin{array}{l} \text { We have, } f(x)=x^{3}-18 x^{2}+96 x \\ \therefore f^{\prime \prime}(x)=3 x^{2}-36 x+96 \\ f^{\prime \prime}(x)=0 \\ \therefore \quad 3 x^{2}-36 x+96=0 \\ \Rightarrow \quad 3\left(x^{2}-12 x+32\right)=0 \\ \Rightarrow \quad(x-8)(x-4)=0 \\ \Rightarrow \quad x=4,8 \\ \end{array}$

For least value of $f(x)$ in 10.91 , we should find the value of function at $x=0,4,8,9$

$\begin{array}{l} f(0)=0 \\ f(4)=4^{3}-18 \times 4^{2}+96 \times 4=64-288+384=160 \\ f(8)=8^{3}-18 \times 8^{2}+96 \times 8=128 \\ f(9)=9^{3}-18 \times 9^{2}+96 \times 9=729-1458+864=13 \end{array}$