Maximums, Minimums, and Everything in Between

Consider the graph of y = f(x) below:



The points A, B, C, D, and E are meant to be where the derivative is zero. This particular function has three relative maximums, two relative minimums, an absolute maximum and no absolute minimum. To see this, remember that a point on the graph, (a, f(a)), is a relative maximum if f(a) > f(x) for all x near a. That is, when we are close to the point a, the function is below the line y = f(a), because f(a) is the largest value near x = a. So, in this example, there are relative maximums when x = A, x = C, and x = E. The same definition is true for a relative minimum except that we are looking for the function value at the minimum to be the smallest, so there are relative minimums when x = B and x = D. You can think of a relative minimum/maximum as a local minimum/maximum and an absolute maximum/minimum as a global maximum/minimum. The relative is what is going on in a small neighborhood. The global is what is going on with the entire function.

The function has an absolute maximum when x = C, because the graph never goes above the line y = f(C). However, the function does not have an absolute minimum because as x goes to negative infinity, f(x) also approaches negative infinity.

In the above example, we were considering f(x) for all x. We can change the problem by looking at f(x) on a restricted set. For example, if we were only interested in f(x) on the interval [A, E], then we still have the relative maximums and minimums as described above and the absolute maximum is f(C). However, we now have an absolute minimum at f(B) because for all x in the interval [A, E], f(B) is less than f(x).

This is how we spot extreme values if we have the graph. Many times however, we will be given a formula for f(x) instead and be asked to find the extreme values. Fortunately, we have a way to deal with this situation. First, we need the definition that c is a critical number of f if f'(c) = 0 or f'(c) does not exist. It is important to remember to check for when the derivative does not exist.

If we would like to find the absolute minimums and maximums of a function f(x) on a closed interval [a,b], we have a very simple process:
  • To find all the possibilities for relative maximums and minimums, we have to find the critical numbers.
  • It could be that the functions exteme values are at the endpoints, so we need to check these points too.
  • Therefore, to solve this type of problem, we find all the critical numbers and make a table of the critical numbers and the endpoints.
  • This table includes all the possible extreme values, so we just need to compare them to see which is the largest/smallest.


    • Example 1
    f(x) = 10 + 6x - x 2 on [-4, 4]
    First find f '(x) = 6 - 2x. Note f '(x) is defined everywhere, so we don't have any troubles there. Next, we set f '(x) = 0 = 6 - 2x, so f '(3) = 0. We can then make a table:

    x -4 3 4
    f(x) -30 19 18

    Comparing values, we see that the absolute maximum is 19 which occurs when x = 3 and the absolute minimum is -30 which occurs when x = -4.

    Example 2
    f(x) = |x - 4| on [-1, 5]
    First we notice that there is a corner in the graph when x = 4, so f '(4) is not defined, so 4 is a critical number. Also, the derivative is never zero. So, we make our table:

    x -1 4 5
    f(x) 5 0 1

    Comparing values, we see that the absolute maximum is 5 which occurs when x = -1 and the absolute minimum is 0 which occurs when x = 4.

    The above process will allow us to find the absolute minimums and maximums of our function. However, it does not give us the relative minimums and maximums. Take a moment to look back at the first graph that we considered. Note that the relative maximum at x = A is actually less than the relative minimum at x = D, so we cannot just compare values to decide whether we have a relative maximum or minimum.

    Again, we do know that relative maximums and minimums do occur at our critical numbers, so we can find f '(x) and then make a list of the points c such that f '(c) = 0 or f '(c) does not exist. To determine whether these are relative maximums, relative minimums, or neither, we have two choices:

    First Derivative Test

    For the first derivative test, we determine when the function is increasing/decreasing and use this to determine if our critical numbers are maximums, minimums, or neither.

    Example
    f(x) = (x 2 - 2x) 3 .
    So, f '(x) = 3(x 2 - 2x) 2 (2x - 2)
    First, f '(x) is defined everywhere. Setting f '(x) = 0:
    f '(x) = 3(x(x - 2)) 2 (2x - 2) = 0, we find that f '(0) = f '(2) = f '(1) = 0
    Remembering that f is increasing if f ' > 0 and f is decreasing if f ' < 0, we can make the following number line by looking at f ' values for x values between our critical numbers:



    Therefore, we see that f is decreasing coming into x = 0 and leaving, so there is neither a max nor a min at x = 0. At x = 1, f is decreasing as x comes in to 1 and is increasing as x leaves 1, so there is a relative minimum at x = 1. I think it helps to draw a rough sketch of f at these points. At x = 2, the function continues to increase, so there is neither a max nor a min at x = 2. This is the first derivative test.

    Second Derivative Test
    The second derivative test is basically the same process as the first derivative test. It uses the fact that the second derivative tells us about the concavity of a function. If f ''(x) > 0, then the f is concave up at x. If in addition, f '(x) = 0, this tells us that there is a minimum at x. Again, it helps to draw a rough sketch of the function. As for the first derivative test, we find the critical numbers. We then check the concavity at those points to determine whether they are maximums, minimums, or neither.

    Consider the same example f(x) = (x 2 - 2x) 3 .
    Then f'(x) = 6(x 2 - 2x) 2 (x - 1)
    So, f ''(x) = 6(x 2 - 2x) 2 (1) + (x - 1)12(x 2 -2x)(2x - 2) = 6(x 2 - 2x)[x 2 -2x + 4(x-1) 2] = 6(x 2 - 2x)[x 2 -2x + 4x 2 -8x + 4] = 6(x 2 - 2x)[5x 2 -10x + 4] = 6x(x -2)5(x - 1-1/sqrt(5))(x -1 + 1/sqrt(5))
    The simplification is not necessary, but it will help if the question also asks for concavity.
    Our critical numbers were x = 0,1,2, so we look at f ''(x) at these values:
    f ''(0) = 0 f ''(1) = 6(-1)(5-10 + 4) = -6(-1) = 6 > 0 f ''(2) = 0
    This tells us that f is concave up at x = 1, so there is a relative minimum at x = 1 and at x = 0 and x = 2 we have neither a minimum or a maximum. This concludes the second derivative test.

    If the problem in addition asks for the concavity of the function, we need to find the points where the second derivative is zero or does not exist. In our example, this occurs when x = 0, x = 2, x = 1 + 1/sqrt(5), x = 1 - 1/sqrt(5) by the quadratic formula. We again make a number line, but we look at the second derivative now.



    In this case, the concavity changes at each of the points, so x = 0, x = 2, x = 1 + 1/sqrt(5), x = 1 - 1/sqrt(5) are points of inflection.

    Challenge
    Use what we have just learned about f(x) to sketch a graph of f(x). Then, input f(x) into your graphing calculator and compare your sketch with the graph.
    See the power of Calculus!!!!!