{"id":3247,"date":"2024-06-13T19:21:05","date_gmt":"2024-06-13T19:21:05","guid":{"rendered":"https:\/\/content.one.lumenlearning.com\/calculus1\/?post_type=chapter&p=3247"},"modified":"2024-08-05T02:16:22","modified_gmt":"2024-08-05T02:16:22","slug":"the-mean-value-theorem-learn-it-3","status":"publish","type":"chapter","link":"https:\/\/content.one.lumenlearning.com\/calculus1\/chapter\/the-mean-value-theorem-learn-it-3\/","title":{"raw":"The Mean Value Theorem: Learn It 3","rendered":"The Mean Value Theorem: Learn It 3"},"content":{"raw":"

Corollaries of the Mean Value Theorem<\/h2>\r\n

Let\u2019s now look at three corollaries of the Mean Value Theorem. These results have important consequences, which we use in upcoming sections.<\/p>\r\n

At this point, we know the derivative of any constant function is zero. The Mean Value Theorem allows us to conclude that the converse is also true. In particular, if [latex]f^{\\prime}(x)=0[\/latex] for all [latex]x[\/latex] in some interval [latex]I[\/latex], then [latex]f(x)[\/latex] is constant over that interval. This result may seem intuitively obvious, but it has important implications that are not obvious, and we discuss them shortly.<\/p>\r\n

\r\n

Corollary 1: functions with a derivative of zero<\/h3>\r\n

Let [latex]f[\/latex] be differentiable over an interval [latex]I[\/latex]. If [latex]f^{\\prime}(x)=0[\/latex] for all [latex]x \\in I[\/latex], then [latex]f(x)[\/latex] is constant for all [latex]x \\in I[\/latex].<\/p>\r\n<\/section>\r\n

\r\n

Proof<\/strong><\/p>\r\n

\r\n

Since [latex]f[\/latex] is differentiable over [latex]I[\/latex], [latex]f[\/latex] must be continuous over [latex]I[\/latex]. Suppose [latex]f(x)[\/latex] is not constant for all [latex]x[\/latex] in [latex]I[\/latex]. Then there exist [latex]a,b \\in I[\/latex], where [latex]a \\ne b[\/latex] and [latex]f(a) \\ne f(b)[\/latex]. Choose the notation so that [latex]a<b[\/latex]. Therefore,<\/p>\r\n

[latex]\\dfrac{f(b)-f(a)}{b-a} \\ne 0[\/latex]<\/div>\r\n

 <\/p>\r\n

Since [latex]f[\/latex] is a differentiable function, by the Mean Value Theorem, there exists [latex]c \\in (a,b)[\/latex] such that<\/p>\r\n

[latex]f^{\\prime}(c)=\\dfrac{f(b)-f(a)}{b-a}[\/latex]<\/div>\r\n

 <\/p>\r\n

Therefore, there exists [latex]c \\in I[\/latex] such that [latex]f^{\\prime}(c) \\ne 0[\/latex], which contradicts the assumption that [latex]f^{\\prime}(x)=0[\/latex] for all [latex]x \\in I[\/latex].<\/p>\r\n

[latex]_\\blacksquare[\/latex]<\/p>\r\n<\/section>\r\n

From the example above, it follows that if two functions have the same derivative, they differ by, at most, a constant.<\/p>\r\n

\r\n

Corollary 2: constant difference theorem<\/h3>\r\n

If [latex]f[\/latex] and [latex]g[\/latex] are differentiable over an interval [latex]I[\/latex] and [latex]f^{\\prime}(x)=g^{\\prime}(x)[\/latex] for all [latex]x \\in I[\/latex], then [latex]f(x)=g(x)+C[\/latex] for some constant [latex]C[\/latex].<\/p>\r\n<\/section>\r\n

\r\n

Proof<\/strong><\/p>\r\n

\r\n

Let [latex]h(x)=f(x)-g(x)[\/latex]. Then, [latex]h^{\\prime}(x)=f^{\\prime}(x)-g^{\\prime}(x)=0[\/latex] for all [latex]x \\in I[\/latex]. By Corollary 1, there is a constant [latex]C[\/latex] such that [latex]h(x)=C[\/latex] for all [latex]x \\in I[\/latex]. Therefore, [latex]f(x)=g(x)+C[\/latex] for all [latex]x \\in I[\/latex].<\/p>\r\n

[latex]_\\blacksquare[\/latex]<\/p>\r\n<\/section>\r\n

The third corollary of the Mean Value Theorem discusses when a function is increasing and when it is decreasing.<\/p>\r\n

\r\n

Recall that a function [latex]f[\/latex] is increasing over [latex]I[\/latex] if [latex]f(x_1) < f(x_2)[\/latex] whenever [latex]x_1 < _2[\/latex], whereas [latex]f[\/latex] is decreasing over [latex]I[\/latex] if [latex]f(x_1) > f(x_2)[\/latex] whenever [latex]x_1 < x_2[\/latex].<\/p>\r\n<\/section>\r\n

Using the Mean Value Theorem, we can show that if the derivative of a function is positive, then the function is increasing; if the derivative is negative, then the function is decreasing (Figure 9).<\/p>\r\n\r\n[caption id=\"attachment_3261\" align=\"alignnone\" width=\"731\"]\"A Figure 9. If a function has a positive derivative over some interval [latex]I[\/latex], then the function increases over that interval [latex]I[\/latex]; if the derivative is negative over some interval [latex]I[\/latex], then the function decreases over that interval [latex]I[\/latex].[\/caption]\r\n\r\n

This fact is important because it means that for a given function [latex]f[\/latex], if there exists a function [latex]F[\/latex] such that [latex]F^{\\prime}(x)=f(x)[\/latex]; then, the only other functions that have a derivative equal to [latex]f[\/latex] are [latex]F(x)+C[\/latex] for some constant [latex]C[\/latex]. We discuss this result in more detail later in the chapter.<\/p>\r\n

\r\n

Corollary 3: increasing and decreasing functions<\/h3>\r\n

Let [latex]f[\/latex] be continuous over the closed interval [latex][a,b][\/latex] and differentiable over the open interval [latex](a,b)[\/latex].<\/p>\r\n

    \r\n\t
  1. If [latex]f^{\\prime}(x)>0[\/latex] for all [latex]x \\in (a,b)[\/latex], then [latex]f[\/latex] is an increasing function over [latex][a,b][\/latex].<\/li>\r\n\t
  2. If [latex]f^{\\prime}(x)<0[\/latex] for all [latex]x \\in (a,b)[\/latex], then [latex]f[\/latex] is a decreasing function over [latex][a,b][\/latex].<\/li>\r\n<\/ol>\r\n<\/section>\r\n
    \r\n

    Proof<\/strong><\/p>\r\n

    \r\n

    We will prove 1.; the proof of 2. is similar. Suppose [latex]f[\/latex] is not an increasing function on [latex]I[\/latex]. Then there exist [latex]a[\/latex] and [latex]b[\/latex] in [latex]I[\/latex] such that [latex]a<b[\/latex], but [latex]f(a) \\ge f(b)[\/latex]. Since [latex]f[\/latex] is a differentiable function over [latex]I[\/latex], by the Mean Value Theorem there exists [latex]c \\in (a,b)[\/latex] such that<\/p>\r\n

    [latex]f^{\\prime}(c)=\\dfrac{f(b)-f(a)}{b-a}[\/latex]<\/div>\r\n

     <\/p>\r\n

    Since [latex]f(a) \\ge f(b)[\/latex], we know that [latex]f(b)-f(a) \\le 0[\/latex]. Also, [latex]a<b[\/latex] tells us that [latex]b-a>0[\/latex]. We conclude that<\/p>\r\n

    [latex]f^{\\prime}(c)=\\dfrac{f(b)-f(a)}{b-a} \\le 0[\/latex]<\/div>\r\n

     <\/p>\r\n

    However, [latex]f^{\\prime}(x)>0[\/latex] for all [latex]x \\in I[\/latex]. This is a contradiction, and therefore [latex]f[\/latex] must be an increasing function over [latex]I[\/latex].<\/p>\r\n

    [latex]_\\blacksquare[\/latex]<\/p>\r\n<\/section>","rendered":"

    Corollaries of the Mean Value Theorem<\/h2>\n

    Let\u2019s now look at three corollaries of the Mean Value Theorem. These results have important consequences, which we use in upcoming sections.<\/p>\n

    At this point, we know the derivative of any constant function is zero. The Mean Value Theorem allows us to conclude that the converse is also true. In particular, if [latex]f^{\\prime}(x)=0[\/latex] for all [latex]x[\/latex] in some interval [latex]I[\/latex], then [latex]f(x)[\/latex] is constant over that interval. This result may seem intuitively obvious, but it has important implications that are not obvious, and we discuss them shortly.<\/p>\n

    \n

    Corollary 1: functions with a derivative of zero<\/h3>\n

    Let [latex]f[\/latex] be differentiable over an interval [latex]I[\/latex]. If [latex]f^{\\prime}(x)=0[\/latex] for all [latex]x \\in I[\/latex], then [latex]f(x)[\/latex] is constant for all [latex]x \\in I[\/latex].<\/p>\n<\/section>\n

    \n

    Proof<\/strong><\/p>\n

    \n

    Since [latex]f[\/latex] is differentiable over [latex]I[\/latex], [latex]f[\/latex] must be continuous over [latex]I[\/latex]. Suppose [latex]f(x)[\/latex] is not constant for all [latex]x[\/latex] in [latex]I[\/latex]. Then there exist [latex]a,b \\in I[\/latex], where [latex]a \\ne b[\/latex] and [latex]f(a) \\ne f(b)[\/latex]. Choose the notation so that [latex]a\n

    [latex]\\dfrac{f(b)-f(a)}{b-a} \\ne 0[\/latex]<\/div>\n

     <\/p>\n

    Since [latex]f[\/latex] is a differentiable function, by the Mean Value Theorem, there exists [latex]c \\in (a,b)[\/latex] such that<\/p>\n

    [latex]f^{\\prime}(c)=\\dfrac{f(b)-f(a)}{b-a}[\/latex]<\/div>\n

     <\/p>\n

    Therefore, there exists [latex]c \\in I[\/latex] such that [latex]f^{\\prime}(c) \\ne 0[\/latex], which contradicts the assumption that [latex]f^{\\prime}(x)=0[\/latex] for all [latex]x \\in I[\/latex].<\/p>\n

    [latex]_\\blacksquare[\/latex]<\/p>\n<\/section>\n

    From the example above, it follows that if two functions have the same derivative, they differ by, at most, a constant.<\/p>\n

    \n

    Corollary 2: constant difference theorem<\/h3>\n

    If [latex]f[\/latex] and [latex]g[\/latex] are differentiable over an interval [latex]I[\/latex] and [latex]f^{\\prime}(x)=g^{\\prime}(x)[\/latex] for all [latex]x \\in I[\/latex], then [latex]f(x)=g(x)+C[\/latex] for some constant [latex]C[\/latex].<\/p>\n<\/section>\n

    \n

    Proof<\/strong><\/p>\n

    \n

    Let [latex]h(x)=f(x)-g(x)[\/latex]. Then, [latex]h^{\\prime}(x)=f^{\\prime}(x)-g^{\\prime}(x)=0[\/latex] for all [latex]x \\in I[\/latex]. By Corollary 1, there is a constant [latex]C[\/latex] such that [latex]h(x)=C[\/latex] for all [latex]x \\in I[\/latex]. Therefore, [latex]f(x)=g(x)+C[\/latex] for all [latex]x \\in I[\/latex].<\/p>\n

    [latex]_\\blacksquare[\/latex]<\/p>\n<\/section>\n

    The third corollary of the Mean Value Theorem discusses when a function is increasing and when it is decreasing.<\/p>\n

    \n

    Recall that a function [latex]f[\/latex] is increasing over [latex]I[\/latex] if [latex]f(x_1) < f(x_2)[\/latex] whenever [latex]x_1 < _2[\/latex], whereas [latex]f[\/latex] is decreasing over [latex]I[\/latex] if [latex]f(x_1) > f(x_2)[\/latex] whenever [latex]x_1 < x_2[\/latex].<\/p>\n<\/section>\n

    Using the Mean Value Theorem, we can show that if the derivative of a function is positive, then the function is increasing; if the derivative is negative, then the function is decreasing (Figure 9).<\/p>\n

    \"A
    Figure 9. If a function has a positive derivative over some interval [latex]I[\/latex], then the function increases over that interval [latex]I[\/latex]; if the derivative is negative over some interval [latex]I[\/latex], then the function decreases over that interval [latex]I[\/latex].<\/figcaption><\/figure>\n

    This fact is important because it means that for a given function [latex]f[\/latex], if there exists a function [latex]F[\/latex] such that [latex]F^{\\prime}(x)=f(x)[\/latex]; then, the only other functions that have a derivative equal to [latex]f[\/latex] are [latex]F(x)+C[\/latex] for some constant [latex]C[\/latex]. We discuss this result in more detail later in the chapter.<\/p>\n

    \n

    Corollary 3: increasing and decreasing functions<\/h3>\n

    Let [latex]f[\/latex] be continuous over the closed interval [latex][a,b][\/latex] and differentiable over the open interval [latex](a,b)[\/latex].<\/p>\n

      \n
    1. If [latex]f^{\\prime}(x)>0[\/latex] for all [latex]x \\in (a,b)[\/latex], then [latex]f[\/latex] is an increasing function over [latex][a,b][\/latex].<\/li>\n
    2. If [latex]f^{\\prime}(x)<0[\/latex] for all [latex]x \\in (a,b)[\/latex], then [latex]f[\/latex] is a decreasing function over [latex][a,b][\/latex].<\/li>\n<\/ol>\n<\/section>\n
      \n

      Proof<\/strong><\/p>\n

      \n

      We will prove 1.; the proof of 2. is similar. Suppose [latex]f[\/latex] is not an increasing function on [latex]I[\/latex]. Then there exist [latex]a[\/latex] and [latex]b[\/latex] in [latex]I[\/latex] such that [latex]a\n

      [latex]f^{\\prime}(c)=\\dfrac{f(b)-f(a)}{b-a}[\/latex]<\/div>\n

       <\/p>\n

      Since [latex]f(a) \\ge f(b)[\/latex], we know that [latex]f(b)-f(a) \\le 0[\/latex]. Also, [latex]a0[\/latex]. We conclude that<\/p>\n

      [latex]f^{\\prime}(c)=\\dfrac{f(b)-f(a)}{b-a} \\le 0[\/latex]<\/div>\n

       <\/p>\n

      However, [latex]f^{\\prime}(x)>0[\/latex] for all [latex]x \\in I[\/latex]. This is a contradiction, and therefore [latex]f[\/latex] must be an increasing function over [latex]I[\/latex].<\/p>\n

      [latex]_\\blacksquare[\/latex]<\/p>\n<\/section>\n","protected":false},"author":15,"menu_order":22,"template":"","meta":{"_candela_citation":"[]","pb_show_title":"on","pb_short_title":"","pb_subtitle":"","pb_authors":[],"pb_section_license":""},"chapter-type":[],"contributor":[],"license":[],"part":652,"module-header":"learn_it","content_attributions":[],"internal_book_links":[],"video_content":null,"cc_video_embed_content":{"cc_scripts":"","media_targets":[]},"try_it_collection":null,"_links":{"self":[{"href":"https:\/\/content.one.lumenlearning.com\/calculus1\/wp-json\/pressbooks\/v2\/chapters\/3247"}],"collection":[{"href":"https:\/\/content.one.lumenlearning.com\/calculus1\/wp-json\/pressbooks\/v2\/chapters"}],"about":[{"href":"https:\/\/content.one.lumenlearning.com\/calculus1\/wp-json\/wp\/v2\/types\/chapter"}],"author":[{"embeddable":true,"href":"https:\/\/content.one.lumenlearning.com\/calculus1\/wp-json\/wp\/v2\/users\/15"}],"version-history":[{"count":6,"href":"https:\/\/content.one.lumenlearning.com\/calculus1\/wp-json\/pressbooks\/v2\/chapters\/3247\/revisions"}],"predecessor-version":[{"id":3265,"href":"https:\/\/content.one.lumenlearning.com\/calculus1\/wp-json\/pressbooks\/v2\/chapters\/3247\/revisions\/3265"}],"part":[{"href":"https:\/\/content.one.lumenlearning.com\/calculus1\/wp-json\/pressbooks\/v2\/parts\/652"}],"metadata":[{"href":"https:\/\/content.one.lumenlearning.com\/calculus1\/wp-json\/pressbooks\/v2\/chapters\/3247\/metadata\/"}],"wp:attachment":[{"href":"https:\/\/content.one.lumenlearning.com\/calculus1\/wp-json\/wp\/v2\/media?parent=3247"}],"wp:term":[{"taxonomy":"chapter-type","embeddable":true,"href":"https:\/\/content.one.lumenlearning.com\/calculus1\/wp-json\/pressbooks\/v2\/chapter-type?post=3247"},{"taxonomy":"contributor","embeddable":true,"href":"https:\/\/content.one.lumenlearning.com\/calculus1\/wp-json\/wp\/v2\/contributor?post=3247"},{"taxonomy":"license","embeddable":true,"href":"https:\/\/content.one.lumenlearning.com\/calculus1\/wp-json\/wp\/v2\/license?post=3247"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}