Calculus Question

Question

Consider the function f(x)=x^3e^4x, −1≤𝑥≤1

The absolute maximum value is:

and this occurs at x equals:

The absolute minimum value is:

and this occurs at x equals:

Answer 1

The Extreme Value Theorem guarantees that a continuous function, such as this one, on a closed interval has an absolute min and absolute max on that interval. It further specifies that those extrema will occur at a critical point for the function (f^'(x) = 0 or f^'(x) undefined) or at one of the endpoints.

For this question, we find f(-1), f(1), and y-values corresponding to any places on the interval where f^'(x) = 0:

f(-1) = - 1/e⁴ , f(1) = e⁴

By product rule, f^'(x) = 3x²e^4x + 4x³e^4x = 0

x²e^4x(4x + 3) = 0 ; x = 0 , x = - 3/4

Plugging in will confirm f(-3/4) is the absolute min, while f(1) is the absolute max.