If the absolute maximum value of the function | vedclass

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(B) Let $f(x) = (x^2 - 2x + 7) e^{g(x)}$,where $g(x) = 4x^3 - 12x^2 - 180x + 31$.First,we find the derivative $f'(x)$:$f'(x) = (2x - 2) e^{g(x)} + (x^

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$\alpha = -3$

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(B) Let $f(x) = (x^2 - 2x + 7) e^{g(x)}$,where $g(x) = 4x^3 - 12x^2 - 180x + 31$.First,we find the derivative $f'(x)$:$f'(x) = (2x - 2) e^{g(x)} + (x^2 - 2x + 7) e^{g(x)} \cdot g'(x)$$g'(x) = 12x^2 - 24x - 180 = 12(x^2 - 2x - 15) = 12(x - 5)(x + 3)$.Substituting $g'(x)$ into $f'(x)$:$f'(x) = e^{g(x)} [2(x - 1) + (x^2 - 2x + 7) \cdot 12(x - 5)(x + 3)]$.For $x \in [-3, 0]$,we observe the sign of $f'(x)$:Since $x \in [-3, 0]$,$(x - 5) Thus,$(x - 5)(x + 3) \le 0$.Also,$(x^2 - 2x + 7)$ is always positive as its discriminant $D = 4 - 28 = -24 Therefore,$(x^2 - 2x + 7) \cdot 12(x - 5)(x + 3) \le 0$.For $x \in [-3, 0]$,$2(x - 1)$ is also negative.Since both terms are non-positive,$f'(x) Since $f'(x) Therefore,the absolute maximum value occurs at the left endpoint of the interval,which is $x = -3$.Thus,$\alpha = -3$.

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

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