Math 205 Midterm #1

Name:

Question Points Score 1 10 2 15 3 15 Total: 40

Answer the questions in the spaces provided on the question sheets. If you run out of room for an answer, continue on the back of the page. When a problem asks for an example, you must justify why your example works. Except in part 3a, where you cannot assume the Implicit Function Theorem, you may use any theorem proved in class.

1 1. (a) (5 points) Give an example of a function f : R2 → R such that the partial deriva- tives of f exist everywhere but f is not continuous. Prove that your example works.

xy Solution: Let f(x, y) = x2+y2 and f(0, 0) = 0. Then f is not continuous at 0, 1 since if we let x = r cos θ and y = r sin θ then f(x, y) = 2 sin 2θ, which has no limit as r → 0. On the other hand,

∂f y3 − x2y ∂f x3 − y2x = = , ∂x (x2 + y2)2 ∂y (x2 + y2)2

exist everywhere other than (0, 0). At (0, 0) we have D1f(0, 0) = limx→0 0 = 0, D2f(0, 0) = limy→0 0 = 0, so the partial derivatives exist everywhere.

(b) (5 points) Prove that the function

x3 f(x, y) = f(0, 0) = 0 x2 + y2

has directional derivatives everywhere, but that it is not C1 at (0, 0).

x3 Solution: Let f(x, y) = x2+y2 . Letting (x, y) = (r cos θ, r sin θ) we have that f(x, y) = r cos3 θ, so the directional derivative at (0, 0) in the direction of ~u = 3 (cos θ, sin θ) is cos θ. On the other hand, D1f(0, 0) = 1 and D2f(0, 0) = 0, so 1 D~uf 6= ~u · (1, 0), and f is not C at (0, 0). However, we know that the directional derivatives exist at (0, 0), and they exist everywhere else because the formulas

x2(x2 + 3y2) x2(2xy) D f(x, y) = D f(x, y)| = 1 (x2 + y2)2 2 (x2 + y2)2

are defined eveywhere away from (0, 0).

Page 2 2. Let f : E → Rn be a C1 function defined on an open subset E ⊆ Rn.

(a) (10 points) Prove that if f is one-to-one and df~x is invertible for all ~x ∈ E then there exists a C1 inverse g : f(E) → E.

Solution: Define g : f(E) → E to be the inverse of f as a map of sets; this exists because f maps E bijectively onto f(E). We just need to show that g is C1. Let ~y ∈ f(E) with ~x ∈ E such that f(~x) = ~y. We will show that f is C1 at ~y; as this will hold for all ~y, we will be done.

As df~x is invertible we know by the Inverse Function Theorem that there is a neighborhood of ~x such that f has a local C1 inverse. This inverse will agree with g at each point, so g must also be C1 at ~y, and we are done.

(b) (5 points) Show that the condition on df~x is necessary by giving an example of a one-to-one C1 f which does not have a C1 inverse. Explain why your example works.

Solution: Let n = 1 and f : R → R be the function x 7→ x3. This is one-to-one, and so has a set-theoretic inverse, but this inverse will not be C1 at x = 0, as the derivative of x 7→ x1/3 tends to ∞ as x → 0.

Page 3 3. (a) (5 points) State the Implicit Function Theorem.

Solution: The Implicit Function Theorem states that, given an open subset E ⊆ Rn+m and a C1 function f : E → Rn, if at a point (a, b) ∈ E (with a ∈ Rn m and b ∈ R ) we have A = df(a,b) and Ax is invertible, then there exists a neighborhood W of b in Rm and a C1 function g : W → Rn such that g(b) = a −1 and f(g(y), y) = 0 for all y ∈ W . In this case, dgb = −Ax Ay. n n m n In this statement, Ax(h) := A(h, 0): R → R and Ay(k) := A(0, k): R → R .

(b) (5 points) At which points (x, y) ∈ R2 is the theorem applicable to the function f(x, y) = xy2 − x3 − x?

Solution: The zero set of f is the line x = 0 and the hyperbola y2 − x2 = 1. The theorem to solve for x in terms of y is applicable everywhere on this zero set except for the intersection points of these two: the two points (0, 1) and (1, 0). The theorem to solve for y in terms of x is available on all of the points on the hyperbola which are not on the line. To see this analytically, note that ∂f/∂x = y2 − 3x2 − 1 and ∂f/∂y = 2xy. To solve for x in terms of y we need ∂f/∂x to be nonzero, so we need y2 −3x2 −1 6= 0and x(y2 − x2 − 1) = 0. If y2 − x2 − 1 = 0 and y2 − 3x2 − 1 = 0 then x = 0, so the only points on the hyperbola where ∂f/∂x = 0 are when x = 0, so y = ±1. On the other hand, if x = 0 then if y2 − 3x2 − 1 = 0 then y = ±1. So the only points where we can’t solve for x in terms of y are (0, ±1). On the other hand, to solve for y in terms of x we need ∂f/∂y to be nonzero, so we need 2xy 6= 0. Thus it applies at points of x(y2 − x2 − 1) = 0 where x, y 6= 0, meaning all points just on the hyperbola y2 − x2 = 1 where x 6= 0.

(c) (5 points) Give an example of a function f : R2 → R and a point (a, b) ∈ R2 such that f(a, b) = 0, the conditions of the Implicit Function Theorem do not hold at (a, b) but there is still an open neighborhood E of b ∈ R and a C1 function g : E → R such that f(g(y), y) = 0 for all y ∈ E and g(b) = a. Prove that your example has the desired properties.

Solution: Let f(x, y) = x2 −y2, and let (a, b) = (0, 0). Then ∂f/∂x = 0, so the Implicit Function Theorem does not apply. However, we can define a function g : R → R defined by g(y) = y, which is clearly C1 and satisfies f(g(y), y) = 0, as desired.

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