Question

If the absolute maximum value of the function
f(x)=(x²–2x+7)e(4x³−12x²−180x+31) in the interval [–3, 0] is f(α), then

• A. α = 0
• B. α = –3
• C. α∈ (–1, 0)
• D. α∈ (–3, –1]

Answer 1

The Correct Option is B

To find the absolute maximum value of the function \( f(x) = (x^2 - 2x + 7)e^{4x^3 - 12x^2 - 180x + 31} \) in the interval \([-3, 0]\), we need to evaluate the function at the critical points within this interval, as well as at the endpoints.

Firstly, determine the endpoints:

• Evaluate \( f(-3) \).
• Evaluate \( f(0) \).

Next, we need to find the critical points by differentiating \( f(x) \) and setting the derivative equal to zero.

The critical points occur where \( f'(x) = 0 \). Due to the complexity of the function, the derivative will also be complex. However, we recognize that such exponential functions often have rapid increases or decreases, particularly due to the cubic polynomial in the exponent.

To evaluate critical points accurately, observe the behavior of the function:

• The component \( e^{4x^3 - 12x^2 - 180x + 31} \) will determine the rapid growth or decay of the function.
• Assess the behavior of \( f(x) \) near the endpoints and identify if it's increasing or decreasing.

For such functions, examining at endpoints usually provides the simplest determination for maximum values. Calculating the values:

• \( f(-3) = ((-3)^2 - 2(-3) + 7)e^{4(-3)^3 - 12(-3)^2 - 180(-3) + 31} = (9 + 6 + 7)e^{-108 - 108 + 540 + 31} \).
• Simplify: \( 22 \times e^{355} \) (When simplified mathematically, this value becomes significantly large due to the exponential term).
• \( f(0) = (0^2 - 2(0) + 7)e^{4(0)^3 - 12(0)^2 - 180(0) + 31} = 7e^{31} \).

Since exponential values increase rapidly with positive exponents, \( f(-3) \), corresponds to a point where the exponential factor is maximizing at its greatest rate in \([-3, 0]\).

Thus, by comparing these calculations and insights about function behavior, we find that:

• The absolute maximum value is at \( x = -3 \).

Hence, the correct answer is α = -3.