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# 14.5 Inverse Function Theorems

14.38   Lemma. Let be a strictly increasing continuous function whose domain is an interval . Then the image of is the interval , and the function has an inverse.

Proof: It is clear that and are in image. Since is continuous we can apply the intermediate value property to conclude that for every number between and there is a number such that , i.e. image. Since is increasing on we have whenever , and thus image . It follows that is surjective, and since strictly increasing functions are injective, is bijective. By remark (14.22) has an inverse.

14.39   Exercise. State and prove the analogue of lemma 14.38 for strictly decreasing functions.

14.40   Exercise. Let be a function whose domain is an interval , and whose image is an interval. Does it follow that is continuous?

14.41   Exercise. A Let be a continuous function on a closed bounded interval . Show that the image of is a closed bounded interval .

14.42   Exercise. Let and be non-empty intervals and let be a continuous function such that image().

a) Show that if is strictly increasing, then the inverse function for is also strictly increasing.

b) Show that if is strictly decreasing, then the inverse function for is also strictly decreasing.

14.43   Theorem (Inverse function theorem.) Let be a continuous
strictly increasing function on an interval of positive length, such that for all . Let be the image of and let

be the inverse function for . Then is differentiable on the interior of and
 (14.44)

Remark: If is a nonvertical line joining two points and then the slope of is

The reflection of about the line whose equation is passes through the points and , so the slope of the reflected line is

Thus theorem 14.43 says that the tangent to graph() at the point is obtained by reflecting the tangent to graph() at about the line whose equation is . This is what you should expect from the geometry of the situation.

Proof of theorem 14.43:      The first thing that should be done, is to prove that is continuous. I am going to omit that proof and just assume the continuity of , and then show that is differentiable, and that is given by formula (14.44).

Let be a point in the interior of . then

 (14.45)

(Observe that we have not divided by zero). Let be a sequence in such that . Then (since is assumed to be continuous), and for all (since is injective). Since is differentiable at , it follows that

Since it follows that

It follows that

and the theorem is proved.

Remark: The inverse function theorem also applies to continuous functions on such that for all in interior . Formula (14.44) is valid in this case also.

Remark: Although we have stated the inverse function theorem for functions on intervals of the form , it holds for functions defined on any interval. Let be an interval, and let be a continuous strictly increasing function from to R such that for all in the interior of . Let be a point in the interior of image. Then we can find points and in image such that . Now maps the interval bijectively onto , and since we can apply the inverse function theorem on the interval to conclude that . It is not necessary to remember the formula for . Once we know that is differentiable, we can calculate by using the chain rule, as illustrated by the examples in the next section.

Next: 14.6 Some Derivative Calculations Up: 14. The Inverse Function Previous: 14.4 The Exponential Function   Index
Ray Mayer 2007-09-07