If the absolute maximum value of the function $$f(x)=\left(x^{2}-2 x+7\right) \mathrm{e}^{\left(4 x^{3}-12 x^{2}-180 x+31\right)}$$ in the interval $[-3,0]$ is $f(\alpha)$, then :

Given, $f(x)=\underbrace{\left(x^{2}-2 x+7\right)}_{f_{1}(x)} \underbrace{e^{\left(4 x^{3}-12 x^{2}-180 x+31\right)}}_{f_{2}(x)}$ <br/><br/> $f_{1}(x)=x^{2}-2 x+7$ <br/><br/> $f_{1}^{\prime}(x)=2 x-2$, so $f(x)$ is decreasing in $[-3,0]$ and positive also <br/><br/> $f_{2}(x)=e^{4 x^{3}-12 x^{2}-180 x+31}$ <br/><br/> $f_{2}^{\prime}(x)=e^{4 x^{3}-12 x^{2}-180 x+31} \cdot 12 x^{2}-24 x-180$ <br/><br/> $=12(x-5)(x+3) e^{4 x^{3}-12 x^{2}-180 x+31}$ <br/><br/> So, $f_{2}(x)$ is also decreasing and positive in $\{-3,0\}$ <br/><br/> $\therefore$ absolute maximum value of $f(x)$ occurs at $x=-3$ <br/><br/> $\therefore \quad \alpha=-3$