Maximum and Minimum

Question

Find the maximum and minimum values attained by the given function on the indicated closed interval.

1. f(x) = 3x^2 - 5x^3 [-2,2]

Answer 1

The first step here is to calculate the first and second order derivatives of f(x).

f'(x) = 6x - 15 x^2

f''(x) = 6 - 30x

Next, we need to find the critical points (candidate points for being minimizers/maximizers). These can be found by setting:

f'(x) = 0 => 6x - 15 x^2 = 0

which has two solutions:

x1 = 0, x2 = x2 = 0.4

Note that both solutions are inside the interval [-2,2]. Nevertheless, we don't know yet if these points are maximizers or minimizers. To find this out, we need to figure out the sign of the second-order derivative at these points. We have

f''(0) = 6 > 0, f''(0.4) = - 6 < 0

Therefore, we now know that x1 = 0 is a local minimum with f(0) = 0 and x2 = 0.4 is a local maximum with f(0.4) = 0.16. We can also infer the following:

1. In the interval [-2, 0], the function f(x) is decreasing.
2. In the interval [0,0.4], f(x) is increasing.
3. In the interval [0.4,2.0], f(x) is decreasing again.

As a result, we still need to check the values of f(x) at x = -2 and x = 2, to see if the points x1, x2 are local or absolute minimizers/maximizers. We have:

f(-2) = 52, f(2) = -28

So x = 0 was only a local minimum, as x = 2 with f(2) = -28) is the absolute minimum. In addition, x = 0.4 was only a local maximum, as x = -2 with f(-2) = 52) is the absolute maximum.

Note that although the initial analysis that we did (find critical points and determine local minima and maxima) was still necessary as we had to know in which parts of the interval [-2,2] the function is decreasing or increasing.

I hope this helped. If you have any further questions, please feel free to reach out.