Find the minimum value of the function f(x) = -x^3 + 3ax on the interval 0 ≤ x ≤ 2. assume that a > 0. give the minimum value in terms of a.

for 0 ≤ a < , minimum is -8 + 6a

              for ≤ 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

Thus, the minimum is -8 + 6a when "a" is between 0 (the endpoint) and .  The minimum is 0 when "a" is between 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 = ,  y = 0 is the lowest value

when a = 2,  y = 0 is the lowest value

not only in mathematics, scientific proof also are required in the other field of science. also maths proofs are of particular structure, containing steps and logics. so proving is like doing science, it is science. that's why it is important to prove. 

an inaccurate proof leads to errors, or stating results that are not accurate.  which wil mess the probelm up you dont want this

step-by-step explanation:

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