Question

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

• A. \(\alpha=0\)
• B. \(\alpha=-3\)
• C. \(\alpha \in(-1,0)\)
• D. \(\alpha \in(-3,-1)\)

Answer 1

The Correct Option is B

To solve this problem, we need to find the absolute maximum value of the function \(f(x)=\left(x^2-2x+7\right) e^{\left(4x^3-12x^2-180x+31\right)}\) over the interval \([-3, 0]\). The goal is to determine the value of \(\alpha\) such that \(f(\alpha)\) represents the maximum value. Let's go through the steps to find the solution:

1. First, we analyze the function \(f(x)\). The term \(e^{\left(4x^3-12x^2-180x+31\right)}\) involves an exponential function, which complicates direct differentiation, so we'll consider the behavior of the polynomial in the exponent separately from the quadratic coefficient:

2. The quadratic portion, \(x^2 - 2x + 7\), is positive for all \(x\) (as it is a parabola opening upwards with a minimum value well above zero due to its vertex form, having minimum value at \(x=1\) which is outside the given interval).

3. Now, observe the exponential term: \(4x^3-12x^2-180x+31\). To analyze this polynomial closely, let's evaluate it at specific critical and boundary points within the interval \([-3, 0]\):

   • At \(x = -3\), the expression \(4(-3)^3 - 12(-3)^2 - 180(-3) + 31\) calculates to a very large value due to high power and positive coefficients from negative terms.
   • At \(x = 0\), \(4(0)^3 - 12(0)^2 - 180(0) + 31 = 31\), which is a constant and smaller compared to the magnitude evaluated at the boundary where \(x = -3\).

4. We may observe that \(x = -3\) potentially presents a larger value upon comparison between \(x = -3\) and \(x = 0\), and intermediate calculations give no significant maximum value increase between these specific boundaries because of the growing negative gradient at each step.

5. The behavior of the exponential term dominates given the massive polynomial outcome making the function highest at \(x = -3\). Therefore, \(f(\alpha)\) is maximized at \(\alpha = -3\).

Conclusion: The absolute maximum value of the function in the given interval occurs at \(\alpha = -3\).