Exercises: Chapter 2, Section 6

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

    Define ch2f1.png by ch2f2.png for ch2f3.png . One has ch2f4.png for all ch2f5.png . The determinant condition guarantees that for each ch2f6.png , there is exactly one solution of ch2f7.png , call that solution ch2f8.png .

    Now, for each ch2f9.png , the Implicit function theorem says that there is a function ch2f10.png defined in an open neighborhood of ch2f11.png and such that ch2f12.png and ch2f13.png is differentiable. By the uniqueness of the solutions in the last paragraph, it must be that ch2f14.png for all ch2f15.png in the domain of ch2f16.png . In particular, the various functions ch2f17.png all glue together into a single function defined on all of ch2f18.png and differentiable everywhere.

    By differentiating the relation ch2f19.png , one gets ch2f20.png for ch2f21.png . Note that this is of the same form as the set of equations for ch2f22.png except that the right hand side functions have changed. An explicit formula can be obtained by using Cramer's rule.

  2. Let ch2f23.png be differentiable. For each ch2f24.png defined ch2f25.png by ch2f26.png . Suppose that for each ch2f27.png there is a unique ch2f28.png with ch2f29.png ; let ch2f30.png be this ch2f31.png .

    In this problem, it is assumed that ch2f32.png was meant to be continuously differentiable.

    1. If ch2f33.png for all ch2f34.png , show that ch2f35.png is differentiable and


      Just as in the last problem, the uniqueness condition guarantees that ch2f37.png is the same as the function provided by the implicit function theorem applied to ch2f38.png . In particular, ch2f39.png is differentiable and differentiating this last relation gives ch2f40.png . Solving for ch2f41.png gives the result.

    2. Show that if ch2f42.png , then for some ch2f43.png we have ch2f44.png and ch2f45.png .

      This follows immediately from part (a).

    3. Let ch2f46.png . Find


      Note that ch2f48.png is defined only when ch2f49.png and ch2f50.png 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 ch2f51.png precisely when ch2f52.png and so the hypothesis of part (a) is true and ch2f53.png . Also, ch2f54.png (since both ch2f55.png and ch2f56.png are positive), and so for fixed ch2f57.png , the minimum of ch2f58.png occurs at ch2f59.png by the second derivative test. Now actually, we are not looking for the minimum over all ch2f60.png , but just for those in the interval ch2f61.png . The derivative ch2f62.png for ch2f63.png . Further ch2f64.png precisely when ch2f65.png and there is a unique ch2f66.png where ch2f67.png . For fixed ch2f68.png , ch2f69.png is achieved at ch2f70.png if ch2f71.png , at ch2f72.png if ch2f73.png , and at ch2f74.png if ch2f75.png .

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

      Suppose ch2f76.png . Then we need to maximize ch2f77.png . The derivative of this function is negative throughout the interval; so the maximum occurs at ch2f78.png . The maximum value is ch2f79.png .

      Suppose ch2f80.png . Then ch2f81.png . The derivative of this function is ch2f82.png . This function has no zeros in the interval because ch2f83.png has derivative ch2f84.png which is always negative in the interval and the value of the function is positive at the right end point. Now ch2f85.png on the interval, and so the maximum must occur at the left hand end point. The maximum value is ch2f86.png . In view of the last paragraph, that means that the maximum over the entire interval ch2f87.png occurs at ch2f88.png .

      Suppose ch2f89.png . Then ch2f90.png . 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 ch2f91.png must therefore occur at ch2f92.png , ch2f93.png , and the value of the maximum is ch2f94.png .