Ashley Stibbins 12-4
Section 2-6 Midterm Review
PROPERTY:
The Intermediate Value Theorem: If function f is continuous for
all x in the closed interval [a,b], and y is a number between f(a)
and f(b), then there is a number x=c in (a,b) for which f(c)=y.
OBJECTIVE:
Given an equation for a continuous function f and a value of y
between f(a) and f(b), find a value of x=c between a and b for
which f(c)=y.
EXAMPLES:
For the problem below, fine, approximately, a value of x=c in
the given interval for which f(c) equals the given y-value. Tell
why the function is continuous, then prove that there is a value
of x=c for which f(c) is exactly to the given y-value. Illustrate by
graph.
1. f(x)= (x-3)^4+2, [1,4], y=8
2. Foot Race Problem: Jesse and Kay run the 100-m race. One
minute after the race begins, Jesse is running 20 km/hr and
Kay is running 15 km/hr. Three minutes after the race begins,
Jesse has slowed to 17 km/hr and Kay has speeded up to 19
km/hr. Assume that each runner’s speed is a continuous
function of time. Prove that there is a time between 1 min and
3 min after the race began at which each one is running exactly
the same speed. Is it possible to tell what that speed is? Is it
possible to tell when that speed occurred? Explain.
3.Prove that the cubic equation x3 + x2 – 4 = 0 has a solution
in the interval (1, 2).
1. Solution: f is continuous because it is a polynomial function.
f(1)=18, f(4)=3
Since 8 is between 18 and 3, there is a number x=c between 1 and 4 for
which f(c)=8, Q.E.D. c=1.4349….
2. Solution: Let f(t)=Jesse’s speed – Kay’s speed.
f(1)= 20-15= 5, which is positive
f(3)= 17-19= -2, which is negative.
Since the speeds are assumed to be continuous, f is also continuous, and
the intermediate value theorem applies.
Thus, there is a value of t between 1 and 3 for which f(t)= 0, meaning
that Jesse and Kay are going exactly the same speed at the same time.
The existence of the time tells you neither what that time is, nor what the
speed is. An existence theorem such as the intermediate value theorem
does not tell these things.
3. 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).