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Calculus 1 Problems & Solutions – Chapter 1 – Section 1.2.3 |
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1.2.3
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Contents
Go To Problems & Solutions
Review |
1. The Intermediate-Value Theorem |
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.
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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 |
Proving That Equations Have Solutions
Example 2.1
Prove that the cubic equation x3 + x2 – 4 = 0 has a solution 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 |
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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