For the function f \colon \mathbb{R} \rightarrow \mathbb{R}, \quad f(x) = (4-x)e^x, consider its global maximum. This global maximum occurs at which value of x?

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BlackTom AI · Accepted Answer

We start by examining the function f(x) = (4 - x) e^x on the real line. To locate any global maximum, derivatives are our first tool. Compute the first derivative using the product rule: f'(x) = derivative of (4 - x) times e^x plus (4 - x) times derivative of e^x. This gives f'(x) = (-1) e^x + (4 - x) e^x = (3 - x) e^x. Since e^x > 0 for all real x, the sign of f'(x) is controlled by the linear factor (3 - x). The critical point occurs where f'(x) = 0, which happens at x = 3.

Next, to classify this critical point, compute the second derivative: start from f'(x) = (3 - x) e^x. Differentiate again: f''(x) = derivative of (3 - x) times e^x plus (3 - x) times derivative of e^x = (-1) e^x + (3 - x) e^x = (2 - x) e^x. Since e^x > 0, f''(x) has the same sign as (2 - x). At x = 3, f''(3) = (2 - 3) e^3 = - e^3 < 0, which confirms a local maximum at x = 3. Because the function grows without bound as x → -∞ or x → ∞? Actually e^x grows for large x, but (4 - x) becomes negative for x > 4, so the product still tends to −∞ in the far right; on the far left, e^x → 0 but (4 - x) grows; the product tends to 0 from the positive side. The checked critical point yields the global maximum due to the sign change of f' around x = 3 and the behavior at infinity, making x = 3 the location of the global maximum.

Therefore, the analysis shows the maximum occurs at x = 3, and the maximum value is f(3) = (4 − 3) e^3 = e^3. If you were choosing among options, you would select the option corresponding to x = 3.

For the function f \colon \mathbb{R} \rightarrow \mathbb{R}, \quad f(x) = (4-x)e^x, consider its global maximum. This global maximum occurs at which value of x?

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