4.5 continuous functions and differentiable functions

Continuous and Differentiable Functions

In this section we highlight some important facts about
continuous functions and differentiable functions.

Elementary Functions

Recall that a formula that may be constructed from the
“basic formulas” by applying the +, – , *, / and the
composition operations in finitely many steps is called
an elementary formula.

an elementary formula. Most commonly used formulas
are elementary formulas.

are elementary formulas. These formulas are also the
ones whose derivatives can be computed easily

ones whose derivatives can be computed easily.
But under this definition, one may constructed an
elementary function whose discontinuity is quite
messy.
http://mathoverflow.net/questions/17901/exis
tence-of-antiderivatives-of-nasty-but-
For a heated discussion:
elementary-functions

ones whose derivatives can be computed easily.
But under this definition, one may constructed an
elementary function whose discontinuity is quite
messy. However there are theorems about the
continuous functions and the differentiable functions
which we may apply to elementary functions.

Continuous Functions

There are two important facts about continuous
functions.

functions.
I. Intermediate Value Theorem

functions.
Let f(x) be a continuous function defined over the
closed interval [a, b] such that f(a) < f(b),

functions.

f(b)

f(a)

a b

functions.
let m be any number where f(a) < m < f(b),

f(b)

m

f(a)

a b

functions.
then there exists at least
one c, i.e. one or more, f(b)
where a < c < b,
m

f(a)

a c b

functions.
then there exists at least
where a < c < b,
m
and that f(c) = m.
f(a)

a c b

functions.
then there exists at least other c’s
where a < c < b,
m
and that f(c) = m.
f(a)

a c b

functions.
then there exists at least other c’s
where a < c < b,
m
and that f(c) = m.
We omit the proof here. f(a)
Here is a link to its proof.
http://en.wikipedia.org/wiki/Intermediate a c b
_value_theorem

Remarks
I. Similarly, if f(a) > f(b) then there is some c with
a < c < b such that f(c) = m for f(a) > m > f(b).

Remarks
II. If the condition is f(a) ≤ m ≤ f(b) then the
conclusion is a ≤ c ≤ b.

Remarks
One important application for this theorem is the
existence of roots.
(Bozano’s Theorem) Let y = f(x) be a continuous
function over the closed interval [a, b], then there
exists at least one c where a < c < b with f(c) = 0
if the signs of f(a) and f(b) are different.

Remarks
One important application for this theorem is the
existence of roots.
(Bozano’s Theorem) Let y = f(x) be a continuous
function over the closed interval [a, b], then there
exists at least one c where a < c < b with f(c) = 0
if the signs of f(a) and f(b) are different.
With this theorem, we conclude that f(x) = x3 – 3x2 – 5
has a root between 2 < x < 5, as in 3.6 Example B,
because f(2) and f(5) have opposite signs.

The other important fact about continuous functions
is the existence of extrema over a closed interval.

II. Extrema Theorem for Continuous Functions

Let y = f(x) be a continuous function defined over a
closed interval V = [a, b], then both the absolute max.
and the absolute min. exist in V.

In particular if M is the maximum and m is the minimum,
then m ≤ f(x) ≤ M for every x in the interval [a, b].

So a continuous function defined over a closed
interval V is always bounded.

So a continuous function defined over a closed
interval V is always bounded.
Corollary. Let y = f(x) be an elementary function
defined over a closed interval V = [a, b], then f(x) is
continuous over [a, b], hence both the absolute max.

Differentiable Functions

We state the following important theorems about
differentiable functions without proofs.

Differentiability is a lot stronger condition than
continuity at a point on a graph.

Theorem. If f(x) is differentiable at x = a then it is
continuous at x = a.

Differentiability means the rate of change may be
measured.

This observation leads to Rolle’s Theorem which
gives the existence of a point c where f'(c) = 0.

measured. The rate of change at an extremum x = c
of f(x) must be 0 because it can’t be + (increasing)
or – (decreasing) hence f'(c) = 0.

measured. The rate of change at an extremum x = c
of f(x) must be 0 because it can’t be + (increasing)
or – (decreasing) hence f'(c) = 0.
This observation leads to Rolle’s Theorem which
gives the existence of a point c where f'(c) = 0.

Rolle’s Theorem
Let f(x) be a differentiable function defined over the
closed interval [a, b] with a < b and that f(a) = f(b),

Rolle’s Theorem

f(a)=f(b)

x
a c b x

Rolle’s Theorem
then there is at least one c where a < c < b
such that f'(c) = 0. other c’s
f'(c) = 0

f(a)=f(b)

x
a c b x

Rolle’s Theorem
Proof. Consider the following f'(c) = 0
two cases.
f(a)=f(b)

x
a c b x

Rolle’s Theorem
two cases.
1. The function f(x) is a f(a)=f(b)

constant function, i.e. x
a c b x
f(x) = f(a) = k, then f'(x) = 0

Rolle’s Theorem
two cases.

a c b x
f(x) = f(a) = k, then f'(x) = 0.
Any number c where a < c < b would satisfy f'(c) = 0.

Rolle’s Theorem
two cases.

a c b x
f(x) = f(a) = k, then f'(x) = 0
Any number c where a < c < b would satisfy f'(c) = 0.
2. If the function f(x) is not a constant function,
then there exists an extremum c between a and b, with
f(c) ≠ f(a) and f(c) ≠ (b) and we must have f'(c) = 0.

The average rate of change of y = f(x) from x = a to b
is f(b) – f(a) = Δy = slope of the chord as shown.
Δx
b–a

Avg. rate of change
y = f(x)
= slope of the chord (b, f(b))
Δy
= Δx
Δy
Δx
(a, f(a))

a b

Δx
b–a
The graph y = f(x) is the rotation of the graph of some
function y = g(x) defined over some interval [A, B]

Avg. rate of change
y = g(x) y = f(x)
g(A) = g(B) Δy
= Δx
(A, g(A)) (B, g(B)) Δy
Δx
(a, f(a))

Rotate

A C B a b

Δx
b–a
The graph y = f(x) is the rotation of the graph of some
function y = g(x) defined over some interval [A, B]
with g(A) = g(B) with the rotation taking
(A,g(A)) to (a, f(a)) and (B,g(B)) to (b, f(b)) as shown.
Avg. rate of change
y = g(x) y = f(x)
g(A) = g(B) Δy
= Δx
(A, g(A)) (B, g(B)) Δy
Δx
(a, f(a))

Rotate

A C B a b

If in addition y = g(x) is differentiable over an interval
that contains [A, B], then Rolle’s Theorem implies the
existence of at least one C where A < C < B such that
the tangent line at (C, g(C)) is a horizontal.

y = f(x)
Rolle’s Theorem
y = g(x) (b, f(b))
g(A) = g(B)
(A, g(A)) (B, g(B))
(a, f(a))

There exists a C
where g'(C) = 0 Rotate
A C B a b

the tangent line at (C, g(C)) is a horizontal. Under the
rotation this horizontal line rotates into a tangent line
that is parallel to the chord from (a, f(a)) to (b, f(b))
y = f(x)
Rolle’s Theorem
y = g(x) (b, f(b))
g(A) = g(B)
(A, g(A)) (B, g(B))
(a, f(a))

There exists a C There exists a C where
where g'(C) = 0 Rotate f '(C) = Avg. rate of change
A C B a b

the tangent line at (C, g(C)) is a horizontal. Under the
rotation this horizontal line rotates into a tangent line
that is parallel to the chord from (a, f(a)) to (b, f(b))
which gives us the Mean Value Theorem. y = f(x)
Rolle’s Theorem Mean Value Theorem
y = g(x) (b, f(b))
g(A) = g(B)
(A, g(A)) (B, g(B))
(a, f(a))

There exists a C There exists a C where
where g'(C) = 0 Rotate f '(C) = Avg. rate of change
A C B a b

Mean Value Theorem
Let y = f(x) be a differentiable function over an interval
that contains [a, b], then there is at least one c where
a < c < b such that y = f(x)

f '(c) = f(b) – f(a) . slope = Avg. rate of change (b, f(b))
b–a =
f(b) – f(a)
b–a .

(a, f(a))

a c b
There exists a c where
f(b) – f(a)
f '(c) = Avg. rate of change =
b–a .

Mean Value Theorem
Let y = f(x) be a differentiable function over an interval
that contains [a, b], then there is at least one c where
a < c < b such that y = f(x)

f '(c) = f(b) – f(a) . slope = Avg. rate of change (b, f(b))
b–a =
f(b) – f(a)
b–a .
Remarks
1. The precise condition
for the theorem is that (a, f(a))

f(x) is continuous in [a, b],
and differentiable in (a, b).
The condition above is
stronger but is sufficient a c b
for our purposes. There exists a c where
f(b) – f(a)
b–a .

2. In statistics, the word “Mean” denotes the “average”.
However, the statement of, the “Mean Value” in the
Mean Value Theorem refers to the average value of
the rate–of–change, y = f(x)

slope = Avg. rate of change (b, f(b))
f(b) – f(a)
=
b–a .

(a, f(a))

a c b
f(b) – f(a)
b–a .


not the “mean value” slope = Avg. rate of change (b, f(b))
f(b) – f(a)
or the average value of =
b–a .
the function f(x) over the
interval [a, b].
(a, f(a))

a c b
f(b) – f(a)
b–a .


not the “mean value” slope = Avg. rate of change (b, f(b))
f(b) – f(a)
or the average value of =
b–a .
interval [a, b].
The average value of (a, f(a))

interval [a, b] is defined
by integrals which are
c
our next topics. a b
f(b) – f(a)
b–a .

Example A.
We left Los Angeles at 12:00 PM arrived at San
Francisco at 3:30 PM that same afternoon covering a
distance of 350 miles.

Example A.
distance of 350 miles. Therefore our average rate is
350 / 3½ miles/hr = 100 mph.

Example A.
350 / 3½ miles/hr = 100 mph. By the Mean Value
Theorem, our speedometer must display the speed at
exactly 100mph at some point in time during our trip.

Example A.
Example B. Find the x and y intercepts of the line
that is tangent to the graph y = √2x + 1 and is parallel
to the chord connecting the points with x = 0 and
x = 4. Use a graphing software to draw the graph to
confirm your answers.

Example A.
Example B. Find the x and y intercepts of the line
that is tangent to the graph y = √2x + 1 and is parallel
to the chord connecting the points with x = 0 and
x = 4. Use a graphing software to draw the graph to
confirm your answers.
The end points in question are (0, 1) and (4, 3) so
the chord connecting them has slope ½.

The function y = √2x + 1 is differentiable and satisfies
the condition of the Mean Value Theorem, therefore
there is some x = c where 0 < c < 4 and y'(c) = ½ .

We may solve for c directly.

y = √2x + 1 → y'(x) = 1/√2x + 1

y = √2x + 1 → y'(x) = 1/√2x + 1
Set 1 = 1
√2x + 1 2

y = √2x + 1 → y'(x) = 1/√2x + 1
Set 1 = 1
√2x + 1 2
√2x + 1 = 2

y = √2x + 1 → y'(x) = 1/√2x + 1
Set 1 = 1
√2x + 1 2
√2x + 1 = 2
2x + 1 = 4
x = 3/2

y = √2x + 1 → y'(x) = 1/√2x + 1
Set 1 = 1
√2x + 1 2
√2x + 1 = 2
2x + 1 = 4
x = 3/2
Therefore the tangent line passes through (3/2, 2) and
it has slope ½.

y = √2x + 1 → y'(x) = 1/√2x + 1
Set 1 = 1
√2x + 1 2
√2x + 1 = 2
2x + 1 = 4
x = 3/2
Therefore the tangent line passes through (3/2, 2) and
it has slope ½. So the equation of the tangent line in
question is y = ½ (x – 3/2) + 2 or y = x/2 + 5/4.
Its x intercept is at –5/2 and the y intercept is at 5/4.

4.5 continuous functions and differentiable functions

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