Calculus Of One Real Variable – By Pheng Kim Ving
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1.2.3 |
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1. The
Intermediate-Value Theorem
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Suppose f
is a continuous function on [a,
b]. See Fig. 1.1. Let v be a real number between f(a) and f(b). Intuitively,
since f is continuous, it takes on every number between f(a) and f(b), ie, every intermediate value. Thus,
v is a value of
f, which means that there exists c between a and b such that f(c) = v.
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Fig. 1.1
f(c) = v. |
Theorem 1.1 – The Intermediate-Value Theorem
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If f is continuous on [a, b] and v lies between f(a) and f(b), then there exists c between a and b such that f(c) = v. |
The proof of this theorem needs the following principle.
Dedekind Cut Principle
Let's partition [a, b] into
two sets A and B such that:
i. Each number in [a, b] is in either A
or B but not both, and
ii. Each number in A is less than every number in B.
See Fig. 1.2.
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Fig. 1.2
Dedekind Cut Partition Of [a, b]. |
Proof Of The Intermediate-Value
Theorem
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Fig. 1.3
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EOP
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2. Applications
Of The Intermediate-Value Theorem
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Proving That Equations Have Solutions
Example 2.1
Prove that the cubic equation x3 + x2 – 4 = 0 has a solition in the interval (1, 2).
Note
A graph of y = x3 + x2 – 4 is sketched in Fig. 2.1. We have:
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Fig. 2.1
Graph Of y = x3 + x2 – 4. |
Solution
Let f(x) = x3 + x2 – 4. Since f
is a polynomial, it's continuous on [1, 2]. We have f(1) = 13 + 12 – 4 = –2 <
0 and f(2) =
23 +
22 –
4 = 8 > 0, so that f(1) < 0 < f(2). So by the
intermediate-value theorem there exists x1 in (1, 2) such that f(x1)
= 0. That is, the equation x3 + x2 – 4 = 0 has a solution in the
interval (1, 2).
EOS
Remark that in the solution we don't have to sketch a graph if not asked to.
Determining Signs Of Functions
Example 2.2
Determine the intervals where f(x) = x3 – 9x is positive and where it's negative.
Solution
EOS
Since f has a constant sign on each of the
intervals, to determine its sign on an interval we evaluate its value at a
point
inside that interval. The sign of that value is the sign of f
on that interval. The point chosen inside each interval should be
the one that makes the calculation of the value of f at it as simple as
possible.
Problems & Solutions
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1. Show that the equation x3 + x – 1 = 0 has a solution in the interval (0, 1).
Solution
Let f(x) = x3 + x
– 1. We have f(0) = 03 + 0 – 1 = –1 < 0 and f(1) = 13 + 1 – 1 = 1
> 0, so that f(0) < 0 < f(1). Thus by
the intermediate-value theorem there exists c in (0, 1) such that f(c)
= 0. That is, the equation x3 + x – 1 = 0 has a
solution in the interval (0, 1).
2. Prove that
the function f defined
by f(x) = x3 – 15x + 1 has at least three zeros in [– 4,
4]. (A point x1
is a zero of f if
f(x1) = 0.)
Solution
We have: f(– 4) = –3 < 0, f(0)
= 1 > 0, f(1) = –13 < 0, and f(4) = 5 > 0. Since f
is continuous on [– 4, 0] and f(– 4) < 0
< f(0), the Intermediate-Value Theorem assures us that there
exists at least one number a in (– 4, 0) such that f(a)
= 0,
ie, f has at least one zero in (– 4, 0). Similarly, f
has at least one zero in (0, 1) and at least one zero in (1, 4). Hence, f
has at least three zeros in (– 4, 0) U (0, 1) U
(1, 4), thus at least three zeros in [– 4, 4].
Note
We chose 0 and 1. Any other pair of real numbers s
and t will also work as long as – 4 < s < t
< 4, f(s) > 0, and f(t)
< 0.
The 0-and-1 pair is the simplest.
3. Find the intervals in which the function f defined by f(x) = x2 + 4x + 3 is positive and negative.
Solution
4. Show that
the function f defined by f(x) = x2
– (a + b – 1)x + ab
takes on the value (a + b)/2, where a and b
are
any two real numbers.
Solution
We have f(x) = x2 – ax – bx + x + ab = (x – a)(x – b) + x.
Case a = b: Replacing b with a we have f(x) = (x – a)2 + x and (a
+ b)/2 = a. Thus, f(a) = (a – a)2
+ a = a =
(a + b)/2.
If a > b, we'll just replace [a, b] by [b, a], and we'll reach the same conclusion.
Solution
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