Download 4.2 Mean Value Theorem

Let
8
<
0 if x  0
x + 2 if 0 < x < 1
g(x) =
:
1 if x 1.
g is defined for all x, is not continuous, however g(x) < 3 for all x, but there is
no number d for which g(d) = 3. Consequently, g has no global maximum value. On
the other hand, g(x) g(0) = 0 for all x, so g(0) = 0 is the global minimum value.
Theorem 4.1.2. Suppose that c is an interior point of an interval I, and f (c) is an
extreme value of f on I. If f 0 (c) exists, then f 0 (c) = 0.
Definition 4.1.2. A number c is called critical number of a function f if f 0 (c) = 0 or
if f is not di↵erentiable at c.
Remark 4.1.2. To find the extreme values of a function f on an interval [a, b] we
proceed as follows:
(a) Find all critical numbers of f in the open interval (a, b). These are the numbers
in (a, b) where either f is not di↵erentiable, or all numbers c in (a, b) where f
is di↵erentiable and such that f 0 (c) = 0.
(b) Compute the value of f at all critical numbers and at the end points a and b.
(c) The largest of the numbers computed in (b) is the maximum value of f on [a, b]
and the smallest of these numbers is the minimum value of f on [a, b].
Warning 4.1.2. Let f (x) = x3 for 1  x  1. f 0 (x) = 3x2 and f 0 (0) = 0 so 0 is a
critical number, but f (0) = 0 is not an extreme value of f on [ 1, 1].
For g(x) = x2 , for 0  x  1. g 0 (x) = 2x and g 0 (0) = 0. However, 0 is not a
critical point, since it is not an interior point of [0, 1]. Nonetheless, g(0) = 0  g(x)
for all 0  x  1. So g(0) = 0 is the minimum value of g on [0, 1].
Let h(x) = x3/4 for x in [0, 1]. h0 (x) = 34 x 1/4 for x 6= 0, and h0 (0) does not exist.
But h(0) = 0 is the minimum value for h.
4.2
Mean Value Theorem
Theorem 4.2.1 (Rolle’s Theorem). Let f be a continuous on [a, b] and di↵erentiable
on (a, b). If f (a) = f (b), then there is a number c in (a, b) such that f 0 (c) = 0.
Theorem 4.2.2 (The Mean Value Theorem (MVT)). Let f be continuous on [a, b]
and di↵erentiable on (a, b). Then there exists a number c in (a, b) such that f 0 (c) =
f (b) f (a)
.
b a
Application 4.2.1. (1) Use Rolle’s theorem to prove that the equation cot x = x
has a solution in (0, ⇡/2).
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(2) Suppose f is di↵erentiable at every number in [ 1, 1] and f 0 (1) = f 0 ( 1) = 1
and assume that f ( 1) = f (1) = 0.
Explain why f has at least one zero in ( 1, 1).
Explain why f 0 must have at least two zeros in ( 1, 1).
4.3
Consequence of the Mean Value Theorem
Definition 4.3.1. If f is a function defined on an interval I, then any di↵erentiable
function F on I such that F 0 (x) = f (x) for every interior point x of I is called an
antiderivative of f .
Theorem 4.3.1. (a) Let f be continuous on an interval I. If f 0 (x) exists and
equals 0 for all interior points x, then f is constant on I.
(b) Let f and g be continuous on an interval I. If f 0 (x) and g 0 (x) exist and are
equal for each interior point x of I, then f g is constant on I. In other words,
there is a number C such that f (x) = g(x) + C for all x in I.
Warning 4.3.1. Let f and g be defined on I = [ 1, 0] [ [2, 3] by
⇢
2 if
1x0
f (x) =
x + 2 if 2  x < 3,
g(x) =
⇢
3 if
1x0
x + 4 if 2  x < 3.
Then f 0 (x) = g 0 (x) for each interior point x of I, but there is no C such that
f (x) = g(x) + C for all x in I. The reason for this is that I is not an interval!
Definition 4.3.2. A function f is increasing on an interval I if f (x) < f (y) for all x, y
in I with x < y.
A function f is decreasing on an interval I if f (x) > f (y) for all x, y in I with
x < y.
Theorem 4.3.2. Let f be a continuous function on an interval I and di↵erentiable
at each interior point of I.
(a) If f 0 (x) > 0 at each interior point of I, then f is increasing on I. Moreover, f
is increasing on I if f 0 (x) > 0 except for a finite number of numbers x in I.
(b) If f 0 (x) < 0 at each interior point of I, then f is decreasing on I. Moreover, f
is decreasing on I if f 0 (x) < 0 except for a finite number of numbers x in I.
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