Department of Mathematics

Colorado State University

Math 317: Homework 7

Due: Friday, April 5, 2019
  1. (28.8) Let when is rational and otherwise.

    1. Show that is continuous at .

    2. Show that is discontinuous at all .

    3. Prove that is differentiable at . (It is insufficient to simply claim )

    1. Proof. Let \((x_n)\) be any sequence that converges to 0. Let \(\epsilon > 0\) be arbitrary. Convergence of \((x_n)\) to 0 means that there exists \(N\) so that for all natural numbers \(n > N\), \(\vert x_n \vert < \sqrt \epsilon \) Then \(\vert f(x_n) \vert \le x_n^2 < \epsilon \). So \((f(x_n))\) converges to \(f(0) = 0\), and so \(f\) is continuous at 0.

    2. Proof. First, suppose is rational. Let be a sequence of irrationals converging to , which must exist by density of irrationals (or, if you like, by example, we have ). Then for all , so converges to 0, but .

      On the other hand, if is irrational, let be a sequence of rationals converging to (if you like, is the sequence of decimal approximations to ), then but so converges to . So is not continuous at any .

    3. Proof. The definition of the derivative of at 0 is . Let . Notice that for , we have that , proving that .

  2. (28.14) Suppose is differentiable at .

    1. Prove that .

    2. Prove that .

    1. Proof. Let , so that there exists for which implies that . But with , this means that implies that . This holds for all , so .

    2. Proof. Notice that (a) actually also proves that . We have that

  3. (29.5) Let be a real-valued function which is differentiable on . Prove that if for all , then is a constant function.

    Proof. If , then for all , . But taking limits as of both sides, we realize that . But this means that is constant.

  4. (29.9) Show that for all .

    Proof. Notice that is true if and only if . Let . Then . Notice that . When , and when , .

    Let . Then , as it must be decreasing always until , when it is its minimum value of . On the other hand, when , the function is always increasing from its minimum value of , so as well. So in all cases .

  5. Let be real-valued and differentiable on , and let . Suppose additionally that .

    Prove that .

    Hint. Use the Mean Value Theorem.

    Proof. Let , so there exists so that for all we have that . Let , and consider the interval . By the Mean Value Theorem, there exists an input between and for which . But then as . This holds for all and all , so .