Additional Material: Section 4.4
To be added After the your turn but before the FACTORING POLYNOMIALS on p. 319
in College Algebra
THE INTERMEDIATE VALUE THEOREM
In our search for zeros, we sometimes encounter irrational zeros. For example, the
polynomial
f ( x) = x 5 − x 4 − 1
Descartes’s rule of signs tells us there is exactly one real positive zero. However, the
rational zero test only yields x = ±1 , neither of which are zeros. So if we know there is a
real positive zero and we know it’s not rational, it must be irrational. Notice that
f (1) = −1 and f (2) = 15 . Since polynomial functions are continuous and the function
goes from negative to positive between x = 1 and x = 2 , we expect a zero somewhere in
that interval. Generating a graph with a graphing utility we find that there is a zero
around x = 1.3
15
12.5
10
7.5
5
2.5
1.2
1.4
1.6
1.8
2
Thu Intermediate Value Theorem states is based on the fact that polynomial functions are
continuous.
Intermediate Value Theorem
Let a and b be real numbers such that a < b and f represent a polynomial function. If
f (a) and f (b) have opposite signs, then there is at least one zero between a and b.
y
f(x)
f(b)
a
b
f(a)
zero
x
Example 11 Using the Intermediate Value Theorem to Approximate Real Zeros
Use the intermediate value theorem to find the real positive zero of
5
4
f ( x) = x − x − 1 .
Step 1: Find two consecutive integer values for x that have corresponding function values
opposite in sign.
x
0
1
2
f (x)
-1
-1
15
f (1) = −1 and f (2) = 15
Step 2: Divide the x interval of length 1 into 10 equal subintervals of length 0.1 and find
the corresponding function values.
We
in
We
we
You
any
x
1.0
1.1
1.2
1.3
1.4
1.5
1.6
1.7
1.8
1.9
2.0
f (x)
-1
-0.854
-0.585
-0.143
0.536
1.53
2.93
4.85
7.40
10.7
15
would have stopped when we arrived at 1.4 because the zero must
occur between x = 1.3 and x = 1.4 since the change in sign occurs
the function values.
can do this same procedure, divide the interval into 10 and when
do we find the zero is somewhere between f (1.32) = −0.285 and
f (1.33) = 0.0326 .
could continue doing this and get an approximation to the zero to
accuracy.
A graphing utility allows you to zoom into the intercept and identify the zero as well.
To three significant digits we say that x = 1.32 is an approximation to the positive real
zero of f ( x) = x 5 − x 4 − 1 .
Additional Exercises
These follow right after Exercise #66 on p. 323
Use the Intermediate Value Theorem to approximate the real zero in the indicated
interval. Approximate to two decimal places.
1.
2.
3.
4.
f ( x) = x 4 − 3 x 3 + 4
f ( x) = x 5 − 3x 3 + 1
f ( x) = 7 x 5 − 2 x 2 + 5 x − 1
f ( x) = −2 x 3 + 3x 2 + 6 x − 7
Answers
1.
2.
3.
4.
1.34
0.74
0.22
– 1.64
[1,2]
[0,1]
[0,1]
[−2,−1]