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Sec on 4.2
    The Mean Value Theorem
          V63.0121.011: Calculus I
        Professor Ma hew Leingang
               New York University


               April 6, 2011


.
Announcements

   Quiz 4 on Sec ons 3.3,
   3.4, 3.5, and 3.7 next
   week (April 14/15)
   Quiz 5 on Sec ons
   4.1–4.4 April 28/29
   Final Exam Thursday May
   12, 2:00–3:50pm
Courant Lecture tomorrow
       Persi Diaconis (Stanford)
     “The Search for Randomness”
      (General Audience Lecture)

    Thursday, April 7, 2011, 3:30pm
    Warren Weaver Hall, room 109
         Recep on to follow

Visit http://cims.nyu.edu/ for details
             and to RSVP
Objectives

   Understand and be able
   to explain the statement
   of Rolle’s Theorem.
   Understand and be able
   to explain the statement
   of the Mean Value
   Theorem.
Outline
 Rolle’s Theorem

 The Mean Value Theorem
    Applica ons

 Why the MVT is the MITC
   Func ons with deriva ves that are zero
   MVT and differen ability
Heuristic Motivation for Rolle’s Theorem
        If you bike up a hill, then back down, at some point your eleva on
        was sta onary.




                                                       .
Image credit: SpringSun
Mathematical Statement of Rolle’s
Theorem
 Theorem (Rolle’s Theorem)


   Let f be con nuous on [a, b]
   and differen able on (a, b).
   Suppose f(a) = f(b). Then
   there exists a point c in
   (a, b) such that f′ (c) = 0.   .
                                      a   b
Mathematical Statement of Rolle’s
Theorem
 Theorem (Rolle’s Theorem)

                                          c
   Let f be con nuous on [a, b]
   and differen able on (a, b).
   Suppose f(a) = f(b). Then
   there exists a point c in
   (a, b) such that f′ (c) = 0.   .
                                      a       b
Flowchart proof of Rolle’s Theorem
                                         endpoints
     Let c be
         .              Let d be
                            .                 .        .
                                          are max
   the max pt          the min pt
                                          and min


                                           f is
     is c.an             is d. an.           .
                 yes               yes   constant
   endpoint?           endpoint?
                                         on [a, b]

        no                  no
    ′                   ′                f′ (x) .≡ 0
    f (c) .= 0         f (d) .= 0
                                         on (a, b)
Outline
 Rolle’s Theorem

 The Mean Value Theorem
    Applica ons

 Why the MVT is the MITC
   Func ons with deriva ves that are zero
   MVT and differen ability
Heuristic Motivation for The Mean Value Theorem
        If you drive between points A and B, at some me your speedometer
        reading was the same as your average speed over the drive.




                                                          .
Image credit: ClintJCL
The Mean Value Theorem
 Theorem (The Mean Value Theorem)

   Let f be con nuous on
   [a, b] and differen able on
   (a, b). Then there exists a
   point c in (a, b) such that

       f(b) − f(a)                          b
                   = f′ (c).        .
          b−a                           a
The Mean Value Theorem
 Theorem (The Mean Value Theorem)

   Let f be con nuous on
   [a, b] and differen able on
   (a, b). Then there exists a
   point c in (a, b) such that

       f(b) − f(a)                          b
                   = f′ (c).        .
          b−a                           a
The Mean Value Theorem
 Theorem (The Mean Value Theorem)

   Let f be con nuous on                    c
   [a, b] and differen able on
   (a, b). Then there exists a
   point c in (a, b) such that

       f(b) − f(a)                              b
                   = f′ (c).        .
          b−a                           a
Rolle vs. MVT
                            f(b) − f(a)
           f′ (c) = 0                   = f′ (c)
                               b−a
              c                    c




                                         b
   .                    .
       a            b          a
Rolle vs. MVT
                                                 f(b) − f(a)
             f′ (c) = 0                                      = f′ (c)
                                                    b−a
                c                                       c




                                                              b
     .                                       .
         a            b                             a
 If the x-axis is skewed the pictures look the same.
Proof of the Mean Value Theorem
 Proof.
 The line connec ng (a, f(a)) and (b, f(b)) has equa on

                                  f(b) − f(a)
                  L(x) = f(a) +               (x − a)
                                     b−a
Proof of the Mean Value Theorem
 Proof.
 The line connec ng (a, f(a)) and (b, f(b)) has equa on

                                  f(b) − f(a)
                  L(x) = f(a) +               (x − a)
                                     b−a
 Apply Rolle’s Theorem to the func on
                                            f(b) − f(a)
       g(x) = f(x) − L(x) = f(x) − f(a) −               (x − a).
                                               b−a
Proof of the Mean Value Theorem
 Proof.
 The line connec ng (a, f(a)) and (b, f(b)) has equa on

                                   f(b) − f(a)
                   L(x) = f(a) +               (x − a)
                                      b−a
 Apply Rolle’s Theorem to the func on
                                             f(b) − f(a)
        g(x) = f(x) − L(x) = f(x) − f(a) −               (x − a).
                                                b−a
 Then g is con nuous on [a, b] and differen able on (a, b) since f is.
Proof of the Mean Value Theorem
 Proof.
 The line connec ng (a, f(a)) and (b, f(b)) has equa on

                                   f(b) − f(a)
                   L(x) = f(a) +               (x − a)
                                      b−a
 Apply Rolle’s Theorem to the func on
                                             f(b) − f(a)
        g(x) = f(x) − L(x) = f(x) − f(a) −               (x − a).
                                                b−a
 Then g is con nuous on [a, b] and differen able on (a, b) since f is.
 Also g(a) = 0 and g(b) = 0 (check both).
Proof of the Mean Value Theorem
 Proof.

                                            f(b) − f(a)
          g(x) = f(x) − L(x) = f(x) − f(a) −             (x − a).
                                               b−a
  So by Rolle’s Theorem there exists a point c in (a, b) such that

                                             f(b) − f(a)
                     0 = g′ (c) = f′ (c) −               .
                                                b−a
Using the MVT to count solutions
 Example
 Show that there is a unique solu on to the equa on x3 − x = 100 in the
 interval [4, 5].
Using the MVT to count solutions
 Example
 Show that there is a unique solu on to the equa on x3 − x = 100 in the
 interval [4, 5].

 Solu on
     By the Intermediate Value Theorem, the func on f(x) = x3 − x
     must take the value 100 at some point on c in (4, 5).
Using the MVT to count solutions
 Example
 Show that there is a unique solu on to the equa on x3 − x = 100 in the
 interval [4, 5].

 Solu on
     By the Intermediate Value Theorem, the func on f(x) = x3 − x
     must take the value 100 at some point on c in (4, 5).
     If there were two points c1 and c2 with f(c1 ) = f(c2 ) = 100,
     then somewhere between them would be a point c3 between
     them with f′ (c3 ) = 0.
Using the MVT to count solutions
 Example
 Show that there is a unique solu on to the equa on x3 − x = 100 in the
 interval [4, 5].

 Solu on
     By the Intermediate Value Theorem, the func on f(x) = x3 − x
     must take the value 100 at some point on c in (4, 5).
     If there were two points c1 and c2 with f(c1 ) = f(c2 ) = 100,
     then somewhere between them would be a point c3 between
     them with f′ (c3 ) = 0.
     However, f′ (x) = 3x2 − 1, which is posi ve all along (4, 5). So
     this is impossible.
Using the MVT to estimate
 Example
 We know that |sin x| ≤ 1 for all x, and that sin x ≈ x for small x.
 Show that |sin x| ≤ |x| for all x.
Using the MVT to estimate
 Example
 We know that |sin x| ≤ 1 for all x, and that sin x ≈ x for small x.
 Show that |sin x| ≤ |x| for all x.

 Solu on
  Apply the MVT to the func on
  f(t) = sin t on [0, x]. We get        Since |cos(c)| ≤ 1, we get
      sin x − sin 0                       sin x
                    = cos(c)                    ≤ 1 =⇒ |sin x| ≤ |x|
          x−0                               x
 for some c in (0, x).
Using the MVT to estimate II


 Example
 Let f be a differen able func on with f(1) = 3 and f′ (x) < 2 for all x
 in [0, 5]. Could f(4) ≥ 9?
Using the MVT to estimate II
 Solu on
    By MVT
          f(4) − f(1)
                      = f′ (c) < 2
             4−1
   for some c in (1, 4). Therefore

   f(4) = f(1) + f′ (c)(3) < 3 + 2 · 3 = 9.

   So no, it is impossible that f(4) ≥ 9.
Using the MVT to estimate II
 Solu on
    By MVT
                                              y       (4, 9)
          f(4) − f(1)
                      = f′ (c) < 2                      (4, f(4))
             4−1
   for some c in (1, 4). Therefore

   f(4) = f(1) + f′ (c)(3) < 3 + 2 · 3 = 9.           (1, 3)
   So no, it is impossible that f(4) ≥ 9.         .            x
Food for Thought
 Ques on
 A driver travels along the New Jersey Turnpike using E-ZPass. The
 system takes note of the me and place the driver enters and exits
 the Turnpike. A week a er his trip, the driver gets a speeding cket
 in the mail. Which of the following best describes the situa on?
 (a) E-ZPass cannot prove that the driver was speeding
 (b) E-ZPass can prove that the driver was speeding
 (c) The driver’s actual maximum speed exceeds his cketed speed
 (d) Both (b) and (c).
Food for Thought
 Ques on
 A driver travels along the New Jersey Turnpike using E-ZPass. The
 system takes note of the me and place the driver enters and exits
 the Turnpike. A week a er his trip, the driver gets a speeding cket
 in the mail. Which of the following best describes the situa on?
 (a) E-ZPass cannot prove that the driver was speeding
 (b) E-ZPass can prove that the driver was speeding
 (c) The driver’s actual maximum speed exceeds his cketed speed
 (d) Both (b) and (c).
Outline
 Rolle’s Theorem

 The Mean Value Theorem
    Applica ons

 Why the MVT is the MITC
   Func ons with deriva ves that are zero
   MVT and differen ability
Functions with derivatives that are zero

 Fact
 If f is constant on (a, b), then f′ (x) = 0 on (a, b).
Functions with derivatives that are zero

 Fact
 If f is constant on (a, b), then f′ (x) = 0 on (a, b).

        The limit of difference quo ents must be 0
        The tangent line to a line is that line, and a constant func on’s
        graph is a horizontal line, which has slope 0.
        Implied by the power rule since c = cx0
Functions with derivatives that are zero

 Ques on
 If f′ (x) = 0 is f necessarily a constant func on?
Functions with derivatives that are zero

 Ques on
 If f′ (x) = 0 is f necessarily a constant func on?

      It seems true
      But so far no theorem (that we have proven) uses informa on
      about the deriva ve of a func on to determine informa on
      about the func on itself
Why the MVT is the MITC
(Most Important Theorem In Calculus!)
  Theorem
  Let f′ = 0 on an interval (a, b).
Why the MVT is the MITC
(Most Important Theorem In Calculus!)
  Theorem
  Let f′ = 0 on an interval (a, b). Then f is constant on (a, b).
Why the MVT is the MITC
(Most Important Theorem In Calculus!)
  Theorem
  Let f′ = 0 on an interval (a, b). Then f is constant on (a, b).

  Proof.
  Pick any points x and y in (a, b) with x < y. Then f is con nuous on
  [x, y] and differen able on (x, y). By MVT there exists a point z in
  (x, y) such that
                          f(y) − f(x)
                                       = f′ (z) = 0.
                             y−x
  So f(y) = f(x). Since this is true for all x and y in (a, b), then f is
  constant.
Functions with the same derivative
 Theorem
 Suppose f and g are two differen able func ons on (a, b) with
 f′ = g′ . Then f and g differ by a constant. That is, there exists a
 constant C such that f(x) = g(x) + C.
Functions with the same derivative
 Theorem
 Suppose f and g are two differen able func ons on (a, b) with
 f′ = g′ . Then f and g differ by a constant. That is, there exists a
 constant C such that f(x) = g(x) + C.

 Proof.
Functions with the same derivative
 Theorem
 Suppose f and g are two differen able func ons on (a, b) with
 f′ = g′ . Then f and g differ by a constant. That is, there exists a
 constant C such that f(x) = g(x) + C.

 Proof.
      Let h(x) = f(x) − g(x)
Functions with the same derivative
 Theorem
 Suppose f and g are two differen able func ons on (a, b) with
 f′ = g′ . Then f and g differ by a constant. That is, there exists a
 constant C such that f(x) = g(x) + C.

 Proof.
      Let h(x) = f(x) − g(x)
      Then h′ (x) = f′ (x) − g′ (x) = 0 on (a, b)
Functions with the same derivative
 Theorem
 Suppose f and g are two differen able func ons on (a, b) with
 f′ = g′ . Then f and g differ by a constant. That is, there exists a
 constant C such that f(x) = g(x) + C.

 Proof.
      Let h(x) = f(x) − g(x)
      Then h′ (x) = f′ (x) − g′ (x) = 0 on (a, b)
      So h(x) = C, a constant
Functions with the same derivative
 Theorem
 Suppose f and g are two differen able func ons on (a, b) with
 f′ = g′ . Then f and g differ by a constant. That is, there exists a
 constant C such that f(x) = g(x) + C.

 Proof.
      Let h(x) = f(x) − g(x)
      Then h′ (x) = f′ (x) − g′ (x) = 0 on (a, b)
      So h(x) = C, a constant
      This means f(x) − g(x) = C on (a, b)
MVT and differentiability
Example
Let
         {
             −x   if x ≤ 0
f(x) =
             x2   if x ≥ 0

Is f differen able at 0?
MVT and differentiability
Example                      Solu on (from the defini on)
Let                          We have
         {
             −x   if x ≤ 0          f(x) − f(0)        −x
f(x) =                          lim−            = lim−    = −1
             x2   if x ≥ 0     x→0     x−0        x→0   x
                                    f(x) − f(0)        x2
Is f differen able at 0?         lim             = lim+ = lim+ x = 0
                               x→0+    x−0        x→0 x    x→0

                             Since these limits disagree, f is not
                             differen able at 0.
MVT and differentiability
Example                      Solu on (Sort of)
Let                          If x < 0, then f′ (x) = −1. If x > 0, then
         {                   f′ (x) = 2x. Since
             −x   if x ≤ 0
f(x) =
             x2   if x ≥ 0        lim+ f′ (x) = 0 and lim− f′ (x) = −1,
                                  x→0                  x→0

Is f differen able at 0?      the limit lim f′ (x) does not exist and so f is
                                       x→0
                             not differen able at 0.
Why only “sort of”?
  This solu on is valid but less           f′ (x)
  direct.                              y   f(x)
  We seem to be using the
  following fact: If lim f′ (x) does
                      x→a
  not exist, then f is not             .   x
  differen able at a.
  equivalently: If f is differen able
  at a, then lim f′ (x) exists.
             x→a
  But this “fact” is not true!
Differentiable with discontinuous derivative
  It is possible for a func on f to be differen able at a even if lim f′ (x)
                                                                 x→a
  does not exist.
  Example
               {
        ′        x2 sin(1/x) if x ̸= 0
  Let f (x) =                          .
                 0            if x = 0
  Then when x ̸= 0,

  f′ (x) = 2x sin(1/x) + x2 cos(1/x)(−1/x2 ) = 2x sin(1/x) − cos(1/x),

  which has no limit at 0.
Differentiable with discontinuous derivative
  It is possible for a func on f to be differen able at a even if lim f′ (x)
                                                                 x→a
  does not exist.
  Example
               {
        ′        x2 sin(1/x) if x ̸= 0
  Let f (x) =                          .
                 0            if x = 0
  However,

      ′          f(x) − f(0)       x2 sin(1/x)
     f (0) = lim             = lim             = lim x sin(1/x) = 0
             x→0    x−0        x→0       x       x→0

  So f′ (0) = 0. Hence f is differen able for all x, but f′ is not
  con nuous at 0!
Differentiability FAIL
              f(x)                           f′ (x)



                .       x                       .          x




 This func on is differen able   But the deriva ve is not
 at 0.                          con nuous at 0!
MVT to the rescue

 Lemma
 Suppose f is con nuous on [a, b] and lim+ f′ (x) = m. Then
                                        x→a

                                f(x) − f(a)
                         lim+               = m.
                        x→a        x−a
MVT to the rescue
 Proof.
 Choose x near a and greater than a. Then
                            f(x) − f(a)
                                        = f′ (cx )
                               x−a
 for some cx where a < cx < x. As x → a, cx → a as well, so:
                   f(x) − f(a)
            lim+               = lim+ f′ (cx ) = lim+ f′ (x) = m.
           x→a        x−a        x→a             x→a
Using the MVT to find limits

 Theorem
 Suppose
                   lim f′ (x) = m1 and lim+ f′ (x) = m2
                  x→a−                  x→a
 If m1 = m2 , then f is differen able at a. If m1 ̸= m2 , then f is not
 differen able at a.
Using the MVT to find limits
 Proof.
 We know by the lemma that
                           f(x) − f(a)
                      lim−             = lim− f′ (x)
                     x→a      x−a        x→a
                           f(x) − f(a)
                      lim+             = lim+ f′ (x)
                     x→a      x−a        x→a

 The two-sided limit exists if (and only if) the two right-hand sides
 agree.
Summary
  Rolle’s Theorem: under suitable condi ons, func ons must
  have cri cal points.
  Mean Value Theorem: under suitable condi ons, func ons
  must have an instantaneous rate of change equal to the
  average rate of change.
  A func on whose deriva ve is iden cally zero on an interval
  must be constant on that interval.
  E-ZPass is kinder than we realized.

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Lesson 19: The Mean Value Theorem (slides)

  • 1. Sec on 4.2 The Mean Value Theorem V63.0121.011: Calculus I Professor Ma hew Leingang New York University April 6, 2011 .
  • 2. Announcements Quiz 4 on Sec ons 3.3, 3.4, 3.5, and 3.7 next week (April 14/15) Quiz 5 on Sec ons 4.1–4.4 April 28/29 Final Exam Thursday May 12, 2:00–3:50pm
  • 3. Courant Lecture tomorrow Persi Diaconis (Stanford) “The Search for Randomness” (General Audience Lecture) Thursday, April 7, 2011, 3:30pm Warren Weaver Hall, room 109 Recep on to follow Visit http://cims.nyu.edu/ for details and to RSVP
  • 4. Objectives Understand and be able to explain the statement of Rolle’s Theorem. Understand and be able to explain the statement of the Mean Value Theorem.
  • 5. Outline Rolle’s Theorem The Mean Value Theorem Applica ons Why the MVT is the MITC Func ons with deriva ves that are zero MVT and differen ability
  • 6. Heuristic Motivation for Rolle’s Theorem If you bike up a hill, then back down, at some point your eleva on was sta onary. . Image credit: SpringSun
  • 7. Mathematical Statement of Rolle’s Theorem Theorem (Rolle’s Theorem) Let f be con nuous on [a, b] and differen able on (a, b). Suppose f(a) = f(b). Then there exists a point c in (a, b) such that f′ (c) = 0. . a b
  • 8. Mathematical Statement of Rolle’s Theorem Theorem (Rolle’s Theorem) c Let f be con nuous on [a, b] and differen able on (a, b). Suppose f(a) = f(b). Then there exists a point c in (a, b) such that f′ (c) = 0. . a b
  • 9. Flowchart proof of Rolle’s Theorem endpoints Let c be . Let d be . . . are max the max pt the min pt and min f is is c.an is d. an. . yes yes constant endpoint? endpoint? on [a, b] no no ′ ′ f′ (x) .≡ 0 f (c) .= 0 f (d) .= 0 on (a, b)
  • 10. Outline Rolle’s Theorem The Mean Value Theorem Applica ons Why the MVT is the MITC Func ons with deriva ves that are zero MVT and differen ability
  • 11. Heuristic Motivation for The Mean Value Theorem If you drive between points A and B, at some me your speedometer reading was the same as your average speed over the drive. . Image credit: ClintJCL
  • 12. The Mean Value Theorem Theorem (The Mean Value Theorem) Let f be con nuous on [a, b] and differen able on (a, b). Then there exists a point c in (a, b) such that f(b) − f(a) b = f′ (c). . b−a a
  • 13. The Mean Value Theorem Theorem (The Mean Value Theorem) Let f be con nuous on [a, b] and differen able on (a, b). Then there exists a point c in (a, b) such that f(b) − f(a) b = f′ (c). . b−a a
  • 14. The Mean Value Theorem Theorem (The Mean Value Theorem) Let f be con nuous on c [a, b] and differen able on (a, b). Then there exists a point c in (a, b) such that f(b) − f(a) b = f′ (c). . b−a a
  • 15. Rolle vs. MVT f(b) − f(a) f′ (c) = 0 = f′ (c) b−a c c b . . a b a
  • 16. Rolle vs. MVT f(b) − f(a) f′ (c) = 0 = f′ (c) b−a c c b . . a b a If the x-axis is skewed the pictures look the same.
  • 17. Proof of the Mean Value Theorem Proof. The line connec ng (a, f(a)) and (b, f(b)) has equa on f(b) − f(a) L(x) = f(a) + (x − a) b−a
  • 18. Proof of the Mean Value Theorem Proof. The line connec ng (a, f(a)) and (b, f(b)) has equa on f(b) − f(a) L(x) = f(a) + (x − a) b−a Apply Rolle’s Theorem to the func on f(b) − f(a) g(x) = f(x) − L(x) = f(x) − f(a) − (x − a). b−a
  • 19. Proof of the Mean Value Theorem Proof. The line connec ng (a, f(a)) and (b, f(b)) has equa on f(b) − f(a) L(x) = f(a) + (x − a) b−a Apply Rolle’s Theorem to the func on f(b) − f(a) g(x) = f(x) − L(x) = f(x) − f(a) − (x − a). b−a Then g is con nuous on [a, b] and differen able on (a, b) since f is.
  • 20. Proof of the Mean Value Theorem Proof. The line connec ng (a, f(a)) and (b, f(b)) has equa on f(b) − f(a) L(x) = f(a) + (x − a) b−a Apply Rolle’s Theorem to the func on f(b) − f(a) g(x) = f(x) − L(x) = f(x) − f(a) − (x − a). b−a Then g is con nuous on [a, b] and differen able on (a, b) since f is. Also g(a) = 0 and g(b) = 0 (check both).
  • 21. Proof of the Mean Value Theorem Proof. f(b) − f(a) g(x) = f(x) − L(x) = f(x) − f(a) − (x − a). b−a So by Rolle’s Theorem there exists a point c in (a, b) such that f(b) − f(a) 0 = g′ (c) = f′ (c) − . b−a
  • 22. Using the MVT to count solutions Example Show that there is a unique solu on to the equa on x3 − x = 100 in the interval [4, 5].
  • 23. Using the MVT to count solutions Example Show that there is a unique solu on to the equa on x3 − x = 100 in the interval [4, 5]. Solu on By the Intermediate Value Theorem, the func on f(x) = x3 − x must take the value 100 at some point on c in (4, 5).
  • 24. Using the MVT to count solutions Example Show that there is a unique solu on to the equa on x3 − x = 100 in the interval [4, 5]. Solu on By the Intermediate Value Theorem, the func on f(x) = x3 − x must take the value 100 at some point on c in (4, 5). If there were two points c1 and c2 with f(c1 ) = f(c2 ) = 100, then somewhere between them would be a point c3 between them with f′ (c3 ) = 0.
  • 25. Using the MVT to count solutions Example Show that there is a unique solu on to the equa on x3 − x = 100 in the interval [4, 5]. Solu on By the Intermediate Value Theorem, the func on f(x) = x3 − x must take the value 100 at some point on c in (4, 5). If there were two points c1 and c2 with f(c1 ) = f(c2 ) = 100, then somewhere between them would be a point c3 between them with f′ (c3 ) = 0. However, f′ (x) = 3x2 − 1, which is posi ve all along (4, 5). So this is impossible.
  • 26. Using the MVT to estimate Example We know that |sin x| ≤ 1 for all x, and that sin x ≈ x for small x. Show that |sin x| ≤ |x| for all x.
  • 27. Using the MVT to estimate Example We know that |sin x| ≤ 1 for all x, and that sin x ≈ x for small x. Show that |sin x| ≤ |x| for all x. Solu on Apply the MVT to the func on f(t) = sin t on [0, x]. We get Since |cos(c)| ≤ 1, we get sin x − sin 0 sin x = cos(c) ≤ 1 =⇒ |sin x| ≤ |x| x−0 x for some c in (0, x).
  • 28. Using the MVT to estimate II Example Let f be a differen able func on with f(1) = 3 and f′ (x) < 2 for all x in [0, 5]. Could f(4) ≥ 9?
  • 29. Using the MVT to estimate II Solu on By MVT f(4) − f(1) = f′ (c) < 2 4−1 for some c in (1, 4). Therefore f(4) = f(1) + f′ (c)(3) < 3 + 2 · 3 = 9. So no, it is impossible that f(4) ≥ 9.
  • 30. Using the MVT to estimate II Solu on By MVT y (4, 9) f(4) − f(1) = f′ (c) < 2 (4, f(4)) 4−1 for some c in (1, 4). Therefore f(4) = f(1) + f′ (c)(3) < 3 + 2 · 3 = 9. (1, 3) So no, it is impossible that f(4) ≥ 9. . x
  • 31. Food for Thought Ques on A driver travels along the New Jersey Turnpike using E-ZPass. The system takes note of the me and place the driver enters and exits the Turnpike. A week a er his trip, the driver gets a speeding cket in the mail. Which of the following best describes the situa on? (a) E-ZPass cannot prove that the driver was speeding (b) E-ZPass can prove that the driver was speeding (c) The driver’s actual maximum speed exceeds his cketed speed (d) Both (b) and (c).
  • 32. Food for Thought Ques on A driver travels along the New Jersey Turnpike using E-ZPass. The system takes note of the me and place the driver enters and exits the Turnpike. A week a er his trip, the driver gets a speeding cket in the mail. Which of the following best describes the situa on? (a) E-ZPass cannot prove that the driver was speeding (b) E-ZPass can prove that the driver was speeding (c) The driver’s actual maximum speed exceeds his cketed speed (d) Both (b) and (c).
  • 33. Outline Rolle’s Theorem The Mean Value Theorem Applica ons Why the MVT is the MITC Func ons with deriva ves that are zero MVT and differen ability
  • 34. Functions with derivatives that are zero Fact If f is constant on (a, b), then f′ (x) = 0 on (a, b).
  • 35. Functions with derivatives that are zero Fact If f is constant on (a, b), then f′ (x) = 0 on (a, b). The limit of difference quo ents must be 0 The tangent line to a line is that line, and a constant func on’s graph is a horizontal line, which has slope 0. Implied by the power rule since c = cx0
  • 36. Functions with derivatives that are zero Ques on If f′ (x) = 0 is f necessarily a constant func on?
  • 37. Functions with derivatives that are zero Ques on If f′ (x) = 0 is f necessarily a constant func on? It seems true But so far no theorem (that we have proven) uses informa on about the deriva ve of a func on to determine informa on about the func on itself
  • 38. Why the MVT is the MITC (Most Important Theorem In Calculus!) Theorem Let f′ = 0 on an interval (a, b).
  • 39. Why the MVT is the MITC (Most Important Theorem In Calculus!) Theorem Let f′ = 0 on an interval (a, b). Then f is constant on (a, b).
  • 40. Why the MVT is the MITC (Most Important Theorem In Calculus!) Theorem Let f′ = 0 on an interval (a, b). Then f is constant on (a, b). Proof. Pick any points x and y in (a, b) with x < y. Then f is con nuous on [x, y] and differen able on (x, y). By MVT there exists a point z in (x, y) such that f(y) − f(x) = f′ (z) = 0. y−x So f(y) = f(x). Since this is true for all x and y in (a, b), then f is constant.
  • 41. Functions with the same derivative Theorem Suppose f and g are two differen able func ons on (a, b) with f′ = g′ . Then f and g differ by a constant. That is, there exists a constant C such that f(x) = g(x) + C.
  • 42. Functions with the same derivative Theorem Suppose f and g are two differen able func ons on (a, b) with f′ = g′ . Then f and g differ by a constant. That is, there exists a constant C such that f(x) = g(x) + C. Proof.
  • 43. Functions with the same derivative Theorem Suppose f and g are two differen able func ons on (a, b) with f′ = g′ . Then f and g differ by a constant. That is, there exists a constant C such that f(x) = g(x) + C. Proof. Let h(x) = f(x) − g(x)
  • 44. Functions with the same derivative Theorem Suppose f and g are two differen able func ons on (a, b) with f′ = g′ . Then f and g differ by a constant. That is, there exists a constant C such that f(x) = g(x) + C. Proof. Let h(x) = f(x) − g(x) Then h′ (x) = f′ (x) − g′ (x) = 0 on (a, b)
  • 45. Functions with the same derivative Theorem Suppose f and g are two differen able func ons on (a, b) with f′ = g′ . Then f and g differ by a constant. That is, there exists a constant C such that f(x) = g(x) + C. Proof. Let h(x) = f(x) − g(x) Then h′ (x) = f′ (x) − g′ (x) = 0 on (a, b) So h(x) = C, a constant
  • 46. Functions with the same derivative Theorem Suppose f and g are two differen able func ons on (a, b) with f′ = g′ . Then f and g differ by a constant. That is, there exists a constant C such that f(x) = g(x) + C. Proof. Let h(x) = f(x) − g(x) Then h′ (x) = f′ (x) − g′ (x) = 0 on (a, b) So h(x) = C, a constant This means f(x) − g(x) = C on (a, b)
  • 47. MVT and differentiability Example Let { −x if x ≤ 0 f(x) = x2 if x ≥ 0 Is f differen able at 0?
  • 48. MVT and differentiability Example Solu on (from the defini on) Let We have { −x if x ≤ 0 f(x) − f(0) −x f(x) = lim− = lim− = −1 x2 if x ≥ 0 x→0 x−0 x→0 x f(x) − f(0) x2 Is f differen able at 0? lim = lim+ = lim+ x = 0 x→0+ x−0 x→0 x x→0 Since these limits disagree, f is not differen able at 0.
  • 49. MVT and differentiability Example Solu on (Sort of) Let If x < 0, then f′ (x) = −1. If x > 0, then { f′ (x) = 2x. Since −x if x ≤ 0 f(x) = x2 if x ≥ 0 lim+ f′ (x) = 0 and lim− f′ (x) = −1, x→0 x→0 Is f differen able at 0? the limit lim f′ (x) does not exist and so f is x→0 not differen able at 0.
  • 50. Why only “sort of”? This solu on is valid but less f′ (x) direct. y f(x) We seem to be using the following fact: If lim f′ (x) does x→a not exist, then f is not . x differen able at a. equivalently: If f is differen able at a, then lim f′ (x) exists. x→a But this “fact” is not true!
  • 51. Differentiable with discontinuous derivative It is possible for a func on f to be differen able at a even if lim f′ (x) x→a does not exist. Example { ′ x2 sin(1/x) if x ̸= 0 Let f (x) = . 0 if x = 0 Then when x ̸= 0, f′ (x) = 2x sin(1/x) + x2 cos(1/x)(−1/x2 ) = 2x sin(1/x) − cos(1/x), which has no limit at 0.
  • 52. Differentiable with discontinuous derivative It is possible for a func on f to be differen able at a even if lim f′ (x) x→a does not exist. Example { ′ x2 sin(1/x) if x ̸= 0 Let f (x) = . 0 if x = 0 However, ′ f(x) − f(0) x2 sin(1/x) f (0) = lim = lim = lim x sin(1/x) = 0 x→0 x−0 x→0 x x→0 So f′ (0) = 0. Hence f is differen able for all x, but f′ is not con nuous at 0!
  • 53. Differentiability FAIL f(x) f′ (x) . x . x This func on is differen able But the deriva ve is not at 0. con nuous at 0!
  • 54. MVT to the rescue Lemma Suppose f is con nuous on [a, b] and lim+ f′ (x) = m. Then x→a f(x) − f(a) lim+ = m. x→a x−a
  • 55. MVT to the rescue Proof. Choose x near a and greater than a. Then f(x) − f(a) = f′ (cx ) x−a for some cx where a < cx < x. As x → a, cx → a as well, so: f(x) − f(a) lim+ = lim+ f′ (cx ) = lim+ f′ (x) = m. x→a x−a x→a x→a
  • 56. Using the MVT to find limits Theorem Suppose lim f′ (x) = m1 and lim+ f′ (x) = m2 x→a− x→a If m1 = m2 , then f is differen able at a. If m1 ̸= m2 , then f is not differen able at a.
  • 57. Using the MVT to find limits Proof. We know by the lemma that f(x) − f(a) lim− = lim− f′ (x) x→a x−a x→a f(x) − f(a) lim+ = lim+ f′ (x) x→a x−a x→a The two-sided limit exists if (and only if) the two right-hand sides agree.
  • 58. Summary Rolle’s Theorem: under suitable condi ons, func ons must have cri cal points. Mean Value Theorem: under suitable condi ons, func ons must have an instantaneous rate of change equal to the average rate of change. A func on whose deriva ve is iden cally zero on an interval must be constant on that interval. E-ZPass is kinder than we realized.