Calculus Of One Real Variable – By Pheng Kim Ving
Chapter 5: Applications Of The Derivative Part 1 – Section 5.1: The Mean-Value Theorem

 

5.1
The Mean-Value Theorem

 

 

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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(x3). The value f(x2) isn't a maximum of f, but there's a vicinity of x2 over which f(x2) is a

 

Fig. 1.1

 

Local maxima and minima.

 

maximum. We call f(x2) a local maximum  of f. The value f(x3) is the minimum of f; there's a vicinity of x3, which can be
as large as [
a, b], over which f(x3) is a minimum; so we also call f(x3) 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 x2, we can say that there exists h > 0 such that f(x2) is a maximum of f on (x2 – h, x2 + 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 (x2 – h, x2 + h) is the intersection of itself and dom( f ).

 

Definitions 1.1 – Local Maxima And Minima

 

 

 

Relative And Absolute Maximum And Minimum

 

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).

 

Extrema

 

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 x3, and that f(x1),
 
f(x2), and f(x3) are local extrema. We observe that the tangent lines to the graph of f at x1 and x2 are horizontal, which
means that
f '(x1) = f '(x2) = 0. There's no (unique) tangent line at x3; 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. 2.1

 

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

 

Theorem 2.1

 

If f(x1) is a local maximum or a local minimum of f and f is differentiable at x1, then f '(x1) = 0.

 

 

Proof

 

Consequently, f '(x1) = 0. If f(x1) is a local minimum, a similar argument will lead to the same conclusion that f '(x1) = 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
x-coordinate of that point, f '(c) = 0.

 

Fig. 3.1

 

f '(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 x2 – 3x + 1 = 0 has a unique solution in [2, 4].

 

Fig. 4.1

 

The graph of y = f(x) = x2 – 3x + 1 cuts the x-axis at two points, but at
exactly one point in [2, 4].

 

Solution

EOS

 

Remark 4.1

 

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 Mean-Value 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 mean-value theorem,
below.

 

Fig. 5.1

 

 

Theorem 5.1 – The Mean-Value 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 mean-value 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
x-axis. 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 Mean-Value Theorem
Define the function g by:

 

EOP

 

Remarks 5.1

 

i.  If f(a) = f(b), then f '(c) = 0. The mean-value theorem is a generalization of Rolle's theorem. Or, in other words,
    Rolle's theorem is a special case of the mean-value theorem.

 

ii.  The mean-value theorem says that the average rate of  change of f over [a, b] is attained at some point in (a, b).

 

iii.  The mean-value 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
    
c1, c2, and c3 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 mean-value 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.

 

Theorem 6.1

 

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
x1 of I, and let C = f(x1), so that C is a constant. Let x be any point of I different from x1.
Applying the mean-value theorem to
f on [x1, x] if x > x1 or on [x, x1] if x < x1, there exists c between x1 and x, thus in
I, such that ( f(x) – f(x1))/(x – x1) = f '(c). But f '(c) = 0. Hence, f(x) = f(x1) = C.
EOP

 

You may wonder what if we re-start the process and pick a new point x1 different from the old point x1 in the proof. Ok,
let's see. By the proof,
f(new x1) = f(old x1) = C. Well, whatever point x1 we pick, f(x1) is always the same value.

 

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7. Proving An Inequality

 

The following example is another application of the mean-value 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
part of the graph of
y = f(x) for all x in [1, 5] lies in the shaded triangle.

 

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Problems & Solutions

 

1.  Prove that the equation 10 – 5x – x2 = 0 has a unique solution in the interval [–1, 2].

 

Solution

 


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Solution

 

Let x1, x2, …, xn be the n distinct points of I where f vanishes. Since f is differentiable on [x1, x2] and f(x1) = f(x2),
Rolle's theorem assures us that there exists
c1,2 in (x1, x2) such that f '(c1,2) = 0. Since f is differentiable on [x2, x3] and
 
f(x2) = f(x3), Rolle's theorem assures us that there exists c2,3 in (x2, x3) such that f '(c2,3) = 0. And so on, until we reach
the interval [
xn–1, xn]. So, we get at least n – 1 distinct points c1,2, c2,3, …, cn–1,n of I where f ' vanishes.

 

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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:

 

    

 

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4.  Prove that the inequality:

 

     

 

Solution

 

 

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Solution

 

For any x in (–2, 4]  f is differentiable on [–2, x]. So by the mean-value 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.

 

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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] it’s also differentiable on [1, 3]. So by the mean-value theorem there exists a in
     (1, 3) thus in (0, 4) such that:

 

 

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