If the absolute maximum value of the function [f(x)=(x^2-2 x+7) e^(4 x^3-12 x^2-180 x+31)]…

Question

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 :

Accepted Answer

Given, $f(x)=\underbrace{\left(x^{2}-2 x+7ight)}_{f_{1}(x)} \underbrace{e^{\left(4 x^{3}-12 x^{2}-180 x+31ight)}}_{f_{2}(x)}$

$f_{1}(x)=x^{2}-2 x+7$

$f_{1}^{\prime}(x)=2 x-2$, so $f(x)$ is decreasing in $[-3,0]$ and positive also

$f_{2}(x)=e^{4 x^{3}-12 x^{2}-180 x+31}$

$f_{2}^{\prime}(x)=e^{4 x^{3}-12 x^{2}-180 x+31} \cdot 12 x^{2}-24 x-180$

$=12(x-5)(x+3) e^{4 x^{3}-12 x^{2}-180 x+31}$

So, $f_{2}(x)$ is also decreasing and positive in $\{-3,0\}$

$	herefore$ absolute maximum value of $f(x)$ occurs at $x=-3$

$	herefore \quad \alpha=-3$

Solutions · 1