# Exercises: Chapter 2, Section 6

1. Use the implicit function theorem to re-do Problem 2-15(c).

Define by for . One has for all . The determinant condition guarantees that for each , there is exactly one solution of , call that solution .

Now, for each , the Implicit function theorem says that there is a function defined in an open neighborhood of and such that and is differentiable. By the uniqueness of the solutions in the last paragraph, it must be that for all in the domain of . In particular, the various functions all glue together into a single function defined on all of and differentiable everywhere.

By differentiating the relation , one gets for . Note that this is of the same form as the set of equations for except that the right hand side functions have changed. An explicit formula can be obtained by using Cramer's rule.

2. Let be differentiable. For each defined by . Suppose that for each there is a unique with ; let be this .

In this problem, it is assumed that was meant to be continuously differentiable.

1. If for all , show that is differentiable and

Just as in the last problem, the uniqueness condition guarantees that is the same as the function provided by the implicit function theorem applied to . In particular, is differentiable and differentiating this last relation gives . Solving for gives the result.

2. Show that if , then for some we have and .

This follows immediately from part (a).

3. Let . Find

Note that is defined only when and are positive. So we need to go back and show that the earlier parts of the problem generalize to this case; there are no difficulties in doing this.

One has precisely when and so the hypothesis of part (a) is true and . Also, (since both and are positive), and so for fixed , the minimum of occurs at by the second derivative test. Now actually, we are not looking for the minimum over all , but just for those in the interval . The derivative for . Further precisely when and there is a unique where . For fixed , is achieved at if , at if , and at if .

We will find where the maximum of the minimum's are located in each of the three cases.

Suppose . Then we need to maximize . The derivative of this function is negative throughout the interval; so the maximum occurs at . The maximum value is .

Suppose . Then . The derivative of this function is . This function has no zeros in the interval because has derivative which is always negative in the interval and the value of the function is positive at the right end point. Now on the interval, and so the maximum must occur at the left hand end point. The maximum value is . In view of the last paragraph, that means that the maximum over the entire interval occurs at .

Suppose . Then . This is a decreasing function and so the maximum occurs at the left hand endpoint. By the result of the previous paragraph the maximum over the entire interval must therefore occur at , , and the value of the maximum is .