{"id":3232,"date":"2024-06-13T18:55:14","date_gmt":"2024-06-13T18:55:14","guid":{"rendered":"https:\/\/content.one.lumenlearning.com\/calculus1\/?post_type=chapter&p=3232"},"modified":"2024-08-05T12:55:30","modified_gmt":"2024-08-05T12:55:30","slug":"maxima-and-minima-learn-it-3","status":"publish","type":"chapter","link":"https:\/\/content.one.lumenlearning.com\/calculus1\/chapter\/maxima-and-minima-learn-it-3\/","title":{"raw":"Maxima and Minima: Learn It 3","rendered":"Maxima and Minima: Learn It 3"},"content":{"raw":"

Extrema and Critical Points<\/h2>\r\n

Locating Absolute Extrema<\/h3>\r\n

The extreme value theorem states that a continuous function over a closed, bounded interval has an absolute maximum and an absolute minimum. Let's look at Figure 2 again.<\/p>\r\n\r\n[caption id=\"\" align=\"aligncenter\" width=\"923\"]\"This Figure 2. Graphs (a), (b), and (c) show several possibilities for absolute extrema for functions with a domain of [latex](\u2212\\infty ,\\infty )[\/latex]. Graphs (d), (e), and (f) show several possibilities for absolute extrema for functions with a domain that is a bounded interval.[\/caption]\r\n\r\n

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<\/strong>, in which case the absolute extremum is a local extremum. Therefore, by Fermat's Theorem, the point [latex]c[\/latex] at which the local extremum occurs must be a critical point. We summarize this result in the following theorem.<\/p>\r\n

\r\n

location of absolute extrema<\/h3>\r\n

Let [latex]f[\/latex] be a continuous function over a closed, bounded interval [latex]I[\/latex]. The absolute maximum of [latex]f[\/latex] over [latex]I[\/latex] and the absolute minimum of [latex]f[\/latex] over [latex]I[\/latex] must occur at endpoints of [latex]I[\/latex] or at critical points of [latex]f[\/latex] in [latex]I[\/latex].<\/p>\r\n<\/section>\r\n

With this idea in mind, let\u2019s examine a procedure for locating absolute extrema.<\/p>\r\n

\r\n

How to: Locate Absolute Extrema over a Closed Interval<\/strong><\/p>\r\n

Consider a continuous function [latex]f[\/latex] defined over the closed interval [latex][a,b][\/latex].<\/p>\r\n

    \r\n\t
  1. Evaluate the Function at Endpoints<\/strong>: Calculate [latex]f(a)[\/latex] and [latex]f(b)[\/latex], where [latex]f[\/latex] is defined on the closed interval [latex][a,b][\/latex].<\/li>\r\n\t
  2. Identify Critical Points<\/strong>: Find all critical points of [latex]f[\/latex] within the interval [latex][a,b][\/latex] and evaluate [latex]f[\/latex] at these points.<\/li>\r\n\t
  3. Determine Extrema<\/strong>: Compare the values from steps 1 and 2. The largest value is the absolute maximum, and the smallest is the absolute minimum of [latex]f[\/latex] on [latex][a,b][\/latex].<\/li>\r\n<\/ol>\r\n<\/section>\r\n

    Now let\u2019s look at how to use this strategy to find the absolute maximum and absolute minimum values for continuous functions.<\/p>\r\n

    \r\n

    For each of the following functions, find the absolute maximum and absolute minimum over the specified interval and state where those values occur.<\/p>\r\n

      \r\n\t
    1. [latex]f(x)=\u2212x^2+3x-2[\/latex] over [latex][1,3][\/latex].<\/li>\r\n\t
    2. [latex]f(x)=x^2-3x^{\\frac{2}{3}}[\/latex] over [latex][0,2][\/latex].<\/li>\r\n<\/ol>\r\n\r\n[reveal-answer q=\"fs-id1165042068485\"]Show Solution[\/reveal-answer]
      \r\n[hidden-answer a=\"fs-id1165042068485\"]\r\n\r\n
        \r\n\t
      1. Step 1<\/strong>. Evaluate [latex]f[\/latex] at the endpoints [latex]x=1[\/latex] and [latex]x=3[\/latex].
        \r\n
        [latex]f(1)=0[\/latex] and [latex]f(3)=-2[\/latex]<\/div>\r\n

        Step 2<\/strong>. Since [latex]f^{\\prime}(x)=-2x+3[\/latex], [latex]f^{\\prime}[\/latex] is defined for all real numbers [latex]x[\/latex]. Therefore, there are no critical points where the derivative is undefined. It remains to check where [latex]f^{\\prime}(x)=0[\/latex]. Since [latex]f^{\\prime}(x)=-2x+3=0[\/latex] at [latex]x=\\frac{3}{2}[\/latex] and [latex]\\frac{3}{2}[\/latex] is in the interval [latex][1,3][\/latex], [latex]f(\\frac{3}{2})[\/latex] is a candidate for an absolute extremum of [latex]f[\/latex] over [latex][1,3][\/latex]. We evaluate [latex]f(\\frac{3}{2})[\/latex] and find<\/p>\r\n

        [latex]f\\left(\\frac{3}{2}\\right)=\\frac{1}{4}[\/latex]<\/div>\r\n

        Step 3<\/strong>. We set up the following table to compare the values found in steps 1 and 2.<\/p>\r\n\r\n\r\n\r\n\r\n\r\n\r\n\r\n
        [latex]x[\/latex]<\/th>\r\n[latex]f(x)[\/latex]<\/th>\r\nConclusion<\/th>\r\n<\/tr>\r\n<\/thead>\r\n
        0<\/td>\r\n0<\/td>\r\n <\/td>\r\n<\/tr>\r\n
        [latex]\\frac{3}{2}[\/latex]<\/td>\r\n[latex]\\frac{1}{4}[\/latex]<\/td>\r\nAbsolute maximum<\/td>\r\n<\/tr>\r\n
        3<\/td>\r\n-2<\/td>\r\nAbsolute minimum<\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\n

        From the table, we find that the absolute maximum of [latex]f[\/latex] over the interval [latex][1,3][\/latex] is [latex]\\frac{1}{4}[\/latex], and it occurs at [latex]x=\\frac{3}{2}[\/latex]. The absolute minimum of [latex]f[\/latex] over the interval [latex][1,3][\/latex] is -2, and it occurs at [latex]x=3[\/latex] as shown in the following graph.<\/p>\r\n\r\n[caption id=\"\" align=\"aligncenter\" width=\"731\"]\"The Figure 8. This function has both an absolute maximum and an absolute minimum.[\/caption]\r\n<\/li>\r\n\t

      2. Step 1<\/strong>. Evaluate [latex]f[\/latex] at the endpoints [latex]x=0[\/latex] and [latex]x=2[\/latex].
        \r\n
        [latex]f(0)=0[\/latex] and [latex]f(2)=4-3\\sqrt[3]{4}\\approx -0.762[\/latex]<\/div>\r\n

        Step 2<\/strong>. The derivative of [latex]f[\/latex] is given by<\/p>\r\n

        [latex]f^{\\prime}(x)=2x-\\dfrac{2}{x^{\\frac{1}{3}}}=\\dfrac{2x^{\\frac{4}{3}}-2}{x^{\\frac{1}{3}}}[\/latex]<\/div>\r\n

        for [latex]x\\ne 0[\/latex]. The derivative is zero when [latex]2x^{\\frac{4}{3}}-2=0[\/latex], which implies [latex]x=\\pm 1[\/latex]. The derivative is undefined at [latex]x=0[\/latex]. Therefore, the critical points of [latex]f[\/latex] are [latex]x=0,1,-1[\/latex]. The point [latex]x=0[\/latex] is an endpoint, so we already evaluated [latex]f(0)[\/latex] in step 1. The point [latex]x=-1[\/latex] is not in the interval of interest, so we need only evaluate [latex]f(1)[\/latex]. We find that<\/p>\r\n

        [latex]f(1)=-2[\/latex]<\/div>\r\n

        Step 3<\/strong>. We compare the values found in steps 1 and 2, in the following table.<\/p>\r\n\r\n\r\n\r\n\r\n\r\n\r\n\r\n
        [latex]x[\/latex]<\/th>\r\n[latex]f(x)[\/latex]<\/th>\r\nConclusion<\/th>\r\n<\/tr>\r\n<\/thead>\r\n
        0<\/td>\r\n0<\/td>\r\nAbsolute maximum<\/td>\r\n<\/tr>\r\n
        1<\/td>\r\n-2<\/td>\r\nAbsolute minimum<\/td>\r\n<\/tr>\r\n
        2<\/td>\r\n-0.762<\/td>\r\n <\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\n

        We conclude that the absolute maximum of [latex]f[\/latex] over the interval [latex][0,2][\/latex] is zero, and it occurs at [latex]x=0[\/latex]. The absolute minimum is \u22122, and it occurs at [latex]x=1[\/latex] as shown in the following graph.<\/p>\r\n\r\n[caption id=\"\" align=\"aligncenter\" width=\"352\"]\"The Figure 9. This function has an absolute maximum at an endpoint of the interval.[\/caption]\r\n<\/li>\r\n<\/ol>\r\n

        Watch the following video to see the worked solution to the first part of this example.<\/p>\r\n