Question

The sum of the global minimum and global maximum values of the function \[ f(x) = \frac{4}{3}x^3 - 4x \quad \text{in } [0, 2] \text{ is} \]

• A. \(0\)
• B. \(\dfrac{8}{3}\)
• C. \(-\dfrac{8}{3}\)
• D. \(1\)

Hint:

Always check both critical points and endpoints when finding global extrema over a closed interval.

Answer 1

The Correct Option is A

We are given the function: \[ f(x) = \frac{4}{3}x^3 - 4x \] To find the global maximum and minimum in the interval \([0, 2]\), evaluate the derivative: \[ f'(x) = 4x^2 - 4 \] Set \(f'(x) = 0\): \[ 4x^2 - 4 = 0 \Rightarrow x = 1 \] Now evaluate the function at the critical point and endpoints: \[ f(0) = 0,\quad f(1) = \frac{4}{3} - 4 = -\frac{8}{3},\quad f(2) = \frac{32}{3} - 8 = \frac{8}{3} \] So the global minimum is \(-\frac{8}{3}\) and the global maximum is \(\frac{8}{3}\). Their sum is: \[ -\frac{8}{3} + \frac{8}{3} = 0 \]