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2.4 Continuity – Calculus Volume 1

The Intermediate Value Theorem

Functions that are continuous over intervals of the form [a,b], where a and b are real numbers, exhibit many useful properties. Throughout our study of calculus, we will encounter many powerful theorems concerning such functions. The first of these theorems is the Intermediate Value Theorem.

The Intermediate Value Theorem

Let f be continuous over a closed, bounded interval [a,b]. If z is any real number between f(a) and f(b), then there is a number c in [a,b] satisfying f(c)=z. (See

(Figure)

).

A diagram illustrating the intermediate value theorem. There is a generic continuous curved function shown over the interval [a,b]. The points fa. and fb. are marked, and dotted lines are drawn from a, b, fa., and fb. to the points (a, fa.) and (b, fb.). A third point, c, is plotted between a and b. Since the function is continuous, there is a value for fc. along the curve, and a line is drawn from c to (c, fc.) and from (c, fc.) to fc., which is labeled as z on the y axis.
Figure 7. There is a number c in [a,b] that satisfies f(c)=z.

Application of the Intermediate Value Theorem

Show that f(x)=x- cos x has at least one zero.

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Solution

Since f(x)=x-cos x is continuous over (−infty,+infty), it is continuous over any closed interval of the form [a,b]. If you can find an interval [a,b] such that f(a) and f(b) have opposite signs, you can use the Intermediate Value Theorem to conclude there must be a real number c in (a,b) that satisfies f(c)=0. Note that

f(0)=0 - cos (0)=-1<0

and

f(frac{pi}{2})=frac{pi}{2} - cos frac{pi}{2}=frac{pi}{2}>0.

Using the Intermediate Value Theorem, we can see that there must be a real number c in [0,pi/2] that satisfies f(c)=0. Therefore, f(x)=x- cos x has at least one zero.

When Can You Apply the Intermediate Value Theorem?

If f(x) is continuous over [0,2], , f(0)>0, and f(2)>0, can we use the Intermediate Value Theorem to conclude that f(x) has no zeros in the interval [0,2]? Explain.

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Solution

No. The Intermediate Value Theorem only allows us to conclude that we can find a value between f(0) and f(2); it doesn’t allow us to conclude that we can’t find other values. To see this more clearly, consider the function f(x)=(x-1)^2. It satisfies f(0)=1>0, , f(2)=1>0, and f(1)=0.

When Can You Apply the Intermediate Value Theorem?

For f(x)=1/x, , f(-1)=-1<0 and f(1)=1>0. Can we conclude that f(x) has a zero in the interval [-1,1]?

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Solution

No. The function is not continuous over [-1,1]. The Intermediate Value Theorem does not apply here.

Show that f(x)=x^3-x^2-3x+1 has a zero over the interval [0,1].

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Solution

f(0)=1>0, , f(1)=-2<0; f(x) is continuous over [0,1]. It must have a zero on this interval.

Hint

Find f(0) and f(1). Apply the Intermediate Value Theorem.

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