f(x) = 3x⁴ - 2x³ - 6x² + 6x + 1 এর চরমমান নির্ণয় কর।

$\mathrm{f}(\mathrm{x})=3 \mathrm{x}^{4}-2 \mathrm{x}^{3}-6 \mathrm{x}^{2}+6 \mathrm{x}+1 \therefore \mathrm{f}^{\prime}(\mathrm{x})=12 \mathrm{x}^{3}-6 \mathrm{x}^{2}-12 \mathrm{x}+6 \therefore \mathrm{f}^{\prime \prime}(\mathrm{x})=36 \mathrm{x}^{2}-12 \mathrm{x}-12$

চরম মানের জন্য, $\mathrm{f}^{\prime}(\mathrm{x})=0 \Rightarrow 12 \mathrm{x}^{3}-6 \mathrm{x}^{2}-12 \mathrm{x}+6=0 \therefore \mathrm{x}=-1,1, \frac{1}{2}$ [Using Calculator]

$\therefore \mathrm{f}(-1)=3(-1)^{4}-2(-1)^{3}-6(-1)^{2}+6(-1)+1=-6$

$\left[\mathrm{f}^{\prime \prime}(-1)=36(-1)^{2}-12(-1)-12=36>0\right.$; লঘুমান]

$\mathrm{f}(1)=3(1)^{4}-2(1)^{3}-6(1)^{2}+6(1)+1=2\left[f^{\prime \prime}(1)=36.1^{2}-12.1-12=12>0\right.$; লঘুমান]

$\mathrm{f}\left(\frac{1}{2}\right)=3\left(\frac{1}{2}\right)^{4}-2\left(\frac{1}{2}\right)^{3}-6\left(\frac{1}{2}\right)^{2}+6\left(\frac{1}{2}\right)+1=\frac{39}{16}\left[\mathrm{f}^{\prime \prime}\left(\frac{1}{2}\right)=36\left(\frac{1}{2}\right)^{2}-12 \cdot \frac{1}{2}-12=-9<0\right.$; গুরুমান]

$\therefore$ চরম মান সমূহঃ $-6,2, \frac{39}{16}$ [যেখানে, $\frac{39}{16}$ হলো গুরুমান এবং -6 ও 2 হলো লঘুমান]