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Find all critical points for f ( x ) = x 3 1 2 x 2 2 x + 1 .

x = 2 3 , x = 1

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Locating absolute extrema

The extreme value theorem states that a continuous function over a closed, bounded interval has an absolute maximum and an absolute minimum. As shown in [link] , one or both of these absolute extrema could occur at an endpoint. If an absolute extremum does not occur at an endpoint, however, it must occur at an interior point, in which case the absolute extremum is a local extremum. Therefore, by [link] , the point c at which the local extremum occurs must be a critical point. We summarize this result in the following theorem.

Location of absolute extrema

Let f be a continuous function over a closed, bounded interval I . The absolute maximum of f over I and the absolute minimum of f over I must occur at endpoints of I or at critical points of f in I .

With this idea in mind, let’s examine a procedure for locating absolute extrema.

Problem-solving strategy: locating absolute extrema over a closed interval

Consider a continuous function f defined over the closed interval [ a , b ] .

  1. Evaluate f at the endpoints x = a and x = b .
  2. Find all critical points of f that lie over the interval ( a , b ) and evaluate f at those critical points.
  3. Compare all values found in (1) and (2). From [link] , the absolute extrema must occur at endpoints or critical points. Therefore, the largest of these values is the absolute maximum of f . The smallest of these values is the absolute minimum of f .

Now let’s look at how to use this strategy to find the absolute maximum and absolute minimum values for continuous functions.

Locating absolute extrema

For each of the following functions, find the absolute maximum and absolute minimum over the specified interval and state where those values occur.

  1. f ( x ) = x 2 + 3 x 2 over [ 1 , 3 ] .
  2. f ( x ) = x 2 3 x 2 / 3 over [ 0 , 2 ] .
  1. Step 1. Evaluate f at the endpoints x = 1 and x = 3 .
    f ( 1 ) = 0 and f ( 3 ) = −2

    Step 2. Since f ( x ) = −2 x + 3 , f is defined for all real numbers x . Therefore, there are no critical points where the derivative is undefined. It remains to check where f ( x ) = 0 . Since f ( x ) = −2 x + 3 = 0 at x = 3 2 and 3 2 is in the interval [ 1 , 3 ] , f ( 3 2 ) is a candidate for an absolute extremum of f over [ 1 , 3 ] . We evaluate f ( 3 2 ) and find
    f ( 3 2 ) = 1 4 .

    Step 3. We set up the following table to compare the values found in steps 1 and 2.
    x f ( x ) Conclusion
    0 0
    3 2 1 4 Absolute maximum
    3 −2 Absolute minimum

    From the table, we find that the absolute maximum of f over the interval [1, 3] is 1 4 , and it occurs at x = 3 2 . The absolute minimum of f over the interval [1, 3] is −2 , and it occurs at x = 3 as shown in the following graph.
    The function f(x) = – x2 + 3x – 2 is graphed from (1, 0) to (3, −2), with its maximum marked at (3/2, 1/4).
    This function has both an absolute maximum and an absolute minimum.
  2. Step 1. Evaluate f at the endpoints x = 0 and x = 2 .
    f ( 0 ) = 0 and f ( 2 ) = 4 3 4 3 0.762

    Step 2. The derivative of f is given by
    f ( x ) = 2 x 2 x 1 / 3 = 2 x 4 / 3 2 x 1 / 3

    for x 0 . The derivative is zero when 2 x 4 / 3 2 = 0 , which implies x = ± 1 . The derivative is undefined at x = 0 . Therefore, the critical points of f are x = 0 , 1 , −1 . The point x = 0 is an endpoint, so we already evaluated f ( 0 ) in step 1. The point x = −1 is not in the interval of interest, so we need only evaluate f ( 1 ) . We find that
    f ( 1 ) = −2 .

    Step 3. We compare the values found in steps 1 and 2, in the following table.
    x f ( x ) Conclusion
    0 0 Absolute maximum
    1 −2 Absolute minimum
    2 −0.762

    We conclude that the absolute maximum of f over the interval [0, 2] is zero, and it occurs at x = 0 . The absolute minimum is −2, and it occurs at x = 1 as shown in the following graph.
    The function f(x) = x2 – 3x2/3 is graphed from (0, 0) to (2, −0.762), with its minimum marked at (1, −2).
    This function has an absolute maximum at an endpoint of the interval.
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Source:  OpenStax, Calculus volume 1. OpenStax CNX. Feb 05, 2016 Download for free at http://cnx.org/content/col11964/1.2
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