Answer: for 0 ≤ a < [tex]\bold{\dfrac{4}{3}}[/tex], minimum is -8 + 6a
for [tex]\bold{\dfrac{4}{3}}[/tex] ≤ a ≤ 2, minimum is 0
Step-by-step explanation:
f(x) = -x³ + 3ax is continuous and differentiable on the interval [0, 2], so the minimum will occur at a critical point or an endpoint.
Step 1: Find the critical point(s) between [0, 2] by setting the derivative of the function equal to zero and solving for x.
f(x) = -x³ + 3ax
f'(x) = -3x² + 3a
0 = -3x² + 3a
3x² = 3a
x² = a
x = ±√a
x = √a and x = -√a
↓
not valid because it is not in the interval [0, 2]
So, the only critical point is at x = √a
Step 2: Evaluate f(x) at the critical point (x = √a) and endpoints (x = 0 and x = 2)
f(x) = -x³ + 3ax
f(0) = -(0)³ + 3a(0)
= 0 + 0
= 0
f(√a) = -(√a)³ + 3a(√a)
= -a√a + 3a√a
= 2a√a
f(2) = -(2)³ + 3a(2)
= -8 + 6a
Step 3: Evaluate the values of "a" that will determine the minimum
Notice that f(a) > f(2) for all a-values so this can be disregarded for the minimum. So, we need to see when f(2) < f(0)
-8 + 6a < 0
6a < 8
[tex]a < \dfrac{4}{3}[/tex]
Thus, the minimum is -8 + 6a when "a" is between 0 (the endpoint) and [tex] \dfrac{4}{3}[/tex]. The minimum is 0 when "a" is between [tex]\dfrac{4}{3}[/tex] and 2 (the endpoint).
Bonus: See the attached graph to see minimum for different values of "a".
as "a" approaches 0, y = -8 is the lowest value
when a = 1, y = -2 is the lowest value
when a = [tex]\dfrac{4}{3}[/tex], y = 0 is the lowest value
when a = 2, y = 0 is the lowest value
