Calculus
Of One Real Variable
By Pheng Kim Ving 
5.1 
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1. Local And
Absolute Maxima And Minima 
We
have in Section
1.2.2 the definitions of a maximum and a minimum of a function. In Fig.
1.1, the maximum of f is f(a)
and the minimum of f is f(x_{3}). The
value f(x_{2}) isn't a maximum of f, but there's a vicinity of x_{2} over which f(x_{2}) is a

Fig. 1.1 Local maxima and minima.

maximum.
We call f(x_{2}) a local maximum of
f. The
value f(x_{3}) is the minimum of f; there's a vicinity of x_{3}, which can be
as large as [a, b], over which f(x_{3}) is a minimum; so we also call f(x_{3}) a local minimum of
f. The
value f(a) is
the
maximum of f;
there's a vicinity to the right of a, which can be as large as [a, b],
over which f(a) is a
maximum; thus
f(a) is also a local maximum of f.
There's a vicinity to the left of b over which f(b) is a minimum; hence f(b) is a local
minimum of f.
For x_{2}, we can say that there exists h >
0 such that f(x_{2}) is a maximum of f on (x_{2} h, x_{2} + h). However, for a, we can
say that there exists k > 0 such that f(a) is a maximum of f on, well, not (a k, a + k), but just [a, a + k). Now, [a, a +
k) is
the intersection of (a k, a + k) and dom( f ). Note that (x_{2} h, x_{2} + h) is the intersection of itself and dom( f ).
Definitions 1.1 Local Maxima And Minima

A
local maximum or minimum is also called a relative maximum or minimum. A
maximum and minimum are also called
absolute maximum and absolute minimum respectively because
they're the maximum and minimum respectively of
the function over its entire domain. An absolute maximum (or minimum) is also a
local maximum (or minimum). The
converse isn't true: a local maximum (or minimum) isn't necessarily an absolute
maximum (or minimum).
An extremum
is either a maximum or a minimum. Maxima and minima are collectively referred
to as extrema.
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2. Derivatives At
Local Maxima And Minima 
Refer
to Fig. 2.1. Suppose that f is defined on [a, b] and differentiable at every point in (a, b)
except x_{3}, and that f(x_{1}),
f(x_{2}), and f(x_{3}) are local extrema. We observe that the tangent
lines to the graph of f at x_{1} and x_{2} are horizontal, which
means that f '(x_{1}) = f '(x_{2}) = 0.
There's no (unique) tangent line at x_{3}; the graph has a sharp point
there. It appears that
the derivatives of f at all points generating local extrema, if they
exist, are 0. Well, they are, as confirmed by the following
theorem. Note that since f isn't defined to the left of a or to
the right of b, it can't be differentiable at either endpoint a or
b.

Fig. 2.1Derivatives at points generating local
extrema, if they exist, are 0.

If f(x_{1}) is a local maximum or a local minimum of f and
f is
differentiable at x_{1}, then f '(x_{1}) =
0. 
Proof
Consequently,
f '(x_{1}) = 0. If f(x_{1}) is a local minimum, a similar argument will lead
to the same conclusion that f '(x_{1}) = 0.
EOP
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3. Rolle's Theorem 
Let f be
continuous on [a, b] and differentiable on (a, b). See
Fig. 3.1. Suppose f(a) = f(b). We observe that there must
be a point on the graph of f where the tangent line to the graph is horizontal,
which means that, letting c be the
xcoordinate
of that point, f '(c) = 0.

Fig. 3.1f '(c) = 0.

Theorem 3.1  Rolle's
Theorem
If f is continuous on [a, b] and differentiable on (a, b)
and f(a) = f(b),
then there exists c in (a, b) such that
f '(c) = 0. 
Proof
EOP
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4. Proving That An
Equation Has A Unique Solution 
The
following example is an application of Rolle's theorem.
Example 4.1
Prove
that the equation x^{2} 3x + 1 = 0 has a unique solution in [2, 4].

Fig. 4.1The graph of y = f(x) = x^{2} 3x + 1 cuts the xaxis at two points, but at

Solution
EOS
The graph
in Fig. 4.1 doesn't constitute proof ! It simply helps us in doing the proof. The graph
of a function is the
visualization of the behavior of the function. The graph doesn't explain why
the function behaves the way it does.
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5. The MeanValue Theorem

Suppose
f is
continuous on [a, b] and differentiable on (a, b), as shown
in Fig. 5.1. Draw the secant line joining the point
(a, f(a)) to
the point (b, f(b)).
Clearly there's a point c where a < c < b such that the tangent line to the graph of f at
x = c is
parallel to, thus has the same slope as, the secant line. The slope of the
secant line is ( f(b) f(a))/(b a), and
that of the tangent line is f '(c). So
( f(b) f(a))/(b a) = f '(c). Recall that the quantity ( f(b) f(a))/(b a) is
also the
average or mean rate of change of f over [a, b] and that the quantity f '(c) is
also the instantaneous rate of change of f
at c; see Section
2.3. Consequently, we see that there's a point c in (a, b) such that the average or mean rate of
change of f over
[a, b]
equals the instantaneous rate of change of f at c. That's confirmed by the meanvalue theorem,
below.

Fig. 5.1 
Theorem 5.1 The MeanValue Theorem
If the function f
is continuous on [a, b] and
differentiable on (a, b), then there
exists c in (a, b)
such that: that is, the mean (or average)
rate of change of f over [a, b] is equal to the instantaneous rate of change of
f at c. 
Note
Here we
discuss why the function g in the proof of the meanvalue theorem is so
defined. Refer to Fig. 5.1. The fact that
the tangent line is parallel to the secant line reminds us of Rolle's theorem,
where the tangent and secant lines are
horizontal, ie, parallel to each other and to the xaxis. So we consider the quantity that equals the
height (or depth) of the
graph of f above
(or below) the secant line. This quantity is a function g such
that the value g(x) at any point x in [a, b]
equals that height (or depth). Thus g(a) = 0 and g(b) = 0, yielding g(a) = g(b). This and the facts that g is
also
continuous on [a, b] and differentiable on (a, b)
enable us to apply Rolle's theorem to g. Now, g(x) is the difference
between f(x) and s(x),
where s(x) is
the equation of the secant line. The secant line passes thru the point (a, f(a)) and
has slope ( f(b) f(a))/(b a). Consequently, its equation is:
Proof Of The MeanValue Theorem
Define the function g by:
i. If f(a) = f(b),
then f '(c) = 0. The meanvalue theorem is a generalization
of Rolle's theorem. Or, in other words,
Rolle's theorem is a special case of
the meanvalue theorem.
ii. The meanvalue
theorem says that the average rate of change
of f over
[a, b] is
attained at some point in (a, b).
iii. The
meanvalue theorem assures that there's at least one point c such
that ( f(b) f(a))/(b a) = f '(c). It doesn't
claim that the point c is
unique. There may be one or more such points c. For example, in Fig. 5.2, there are 3 points
c_{1}, c_{2}, and c_{3} such that:
interchangeably.

Fig. 5.2

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6. Functions Having Derivative Equal To 0

We saw
in Section
3.1 Theorem 2.1 that if a function is constant, then its derivative is 0.
We're now going to see that
the converse is also true, ie, if the derivative of a function is 0 on an
interval, then the function is constant on that interval.
This result is an application of the meanvalue theorem. Intuitively, this is
obvious: since the derivative or rate of change
of the function is 0, the function doesn't change, ie, it's constant. Or,
geometrically, since the tangent line to the graph of
the function at every point of the graph is horizontal, the graph must be a
horizontal line or line segment, and thus the
function is constant.
Suppose f is
differentiable on an interval I. If f '(x) = 0 for all x in I then f is constant on I. 
Proof
Pick an arbitrary point x_{1} of I, and let C = f(x_{1}), so that C is a
constant. Let x be any point of I different from x_{1}.
Applying the meanvalue theorem to f on [x_{1}, x] if x > x_{1} or on [x, x_{1}] if x < x_{1}, there exists c
between x_{1} and x, thus in
I, such
that ( f(x) f(x_{1}))/(x x_{1}) = f '(c). But f '(c) = 0.
Hence, f(x) = f(x_{1}) = C.
EOP
You may
wonder what if we restart the process and pick a new point x_{1} different from the old point x_{1} in the proof. Ok,
let's see. By the proof, f(new x_{1}) = f(old x_{1}) = C. Well, whatever point x_{1} we pick, f(x_{1}) is always the same value.
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7. Proving An
Inequality 
The
following example is another application of the meanvalue theorem.
Example 7.1
Prove that
the inequality:
holds
true for all x > 0.
Solution
EOS
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8. Bounding Functions 
Example 8.1
Solution
EOS

Fig.
8.1
Bounding functions of f(x) on [1, 5] are y = 2x 1 and y = 2x 5. The

Problems &
Solutions 
1.
Prove that the equation 10 5x
x^{2} = 0 has a unique solution in the
interval [1, 2].
Solution
Solution
Let
x_{1}, x_{2},
, x_{n} be the n distinct points of I
where f vanishes. Since f is differentiable on [x_{1}, x_{2}] and f(x_{1}) = f(x_{2}),
Rolle's theorem assures us that there exists c_{1,2} in (x_{1}, x_{2}) such that f
'(c_{1,2}) = 0. Since f is differentiable on [x_{2}, x_{3}] and
f(x_{2}) = f(x_{3}), Rolle's
theorem assures us that there exists c_{2,3} in (x_{2}, x_{3}) such that f
'(c_{2,3}) = 0. And so
on, until we reach
the interval [x_{n}_{1}, x_{n}]. So, we get at least n
1 distinct points c_{1,2}, c_{2,3},
, c_{n}_{1,}_{n}
of I where f '
vanishes.
3.
For each of the following functions f with the indicated interval of the
form [a, b], find the points c in (a, b) where
the
tangent line to the graph of f
is parallel to the secant line joining
(a, f(a))
and (b, f(b)).
Solution
a.
Let S be the secant line joining (a,
f(a))
and (b,
f(b)).
The slope of S is:
Thus c is in (a,
b)
and the tangent line at x = c is parallel to S.
b.
Let S be the secant line joining (1, f(1))
and (2, f(2)). The slope of S
is:
4. Prove that
the inequality:
Solution
Solution
For
any x
in (2, 4] f is differentiable
on [2, x]. So by the meanvalue theorem there
exists c in (2, x) such that:
The
bounding functions of f(x) on [2, 4] are y
= 5x
9 and y = 5x + 11.
6.
Suppose that a function f
is twice differentiable on [0, 4] and
that f(1) = f(2) =
0 and f(3) = 1. Prove that:
a. f '(a)
= 1/2 for some point a in (0, 4).
b. f ''(b)
> 1/2 for some point b
in (0, 4).
Solution
a.
Because f is differentiable on [0, 4] its also
differentiable on [1, 3]. So by the meanvalue theorem there exists a
in
(1, 3) thus in (0, 4) such that:
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