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Calculus and Analysis in Euclidean Space

Jerry Shurman

Chapter 5

Inverse and Implicit Functions - all with Video Answers

Educators

Section 1

Preliminaries

01:15

Problem 1

Let $x \in B(a ; \varepsilon)$ and let $\delta=\varepsilon-|x-a|$. Explain why $\delta>0$ and why $B(x ; \delta) \subset B(a ; \varepsilon)$.

R M
R M
Numerade Educator
01:02

Problem 2

Show that a subset of $\mathbb{R}^n$ is open if and only if each of its points is interior.

Raj Bala
Raj Bala
Numerade Educator

Problem 3

Prove that every closed ball $\bar{B}$ is indeed a closed set, as is its boundary $\partial \bar{B}$. Show that every closed ball and its boundary are also bounded, hence compact.

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Problem 4

Find a continuous function $f: \mathbb{R}^n \longrightarrow \mathbb{R}^m$ and an open set $A \subset \mathbb{R}^n$ such that the image $f(A) \subset \mathbb{R}^m$ of $A$ under $f$ is not open. Feel free to choose $n$ and $m$.

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03:56

Problem 5

Define $f: \mathbb{R} \longrightarrow \mathbb{R}$ by $f(x)=x^3-3 x$. Compute $f(-1 / 2)$. Find $f^{-1}((0,11 / 8)), f^{-1}((0,2)), f^{-1}((-\infty,-11 / 8) \cup(11 / 8, \infty))$. Does $f^{-1}$ exist?

James Kiss
James Kiss
Numerade Educator

Problem 6

Show that for $f: \mathbb{R}^n \longrightarrow \mathbb{R}^m$ and $B \subset \mathbb{R}^m$, the inverse image of the complement is the complement of the inverse image,

$$
f^{-1}\left(B^c\right)=f^{-1}(B)^c
$$

Does the analogous formula hold for forward images?

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Problem 7

If $f: \mathbb{R}^n \longrightarrow \mathbb{R}^m$ is continuous and $B \subset \mathbb{R}^m$ is closed, show that $f^{-1}(B)$ is closed. What does this say about the level sets of continuous functions?

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Problem 8

Prove the converse to Theorem 5.1.2: if $f: A \longrightarrow \mathbb{R}^m$ (where $A \subset \mathbb{R}^n$ is open) is such that for every open $W \subset \mathbb{R}^m$ also $f^{-1}(W) \subset A$ is open, then $f$ is continuous.

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03:38

Problem 9

Let $a$ and $b$ be real numbers with $a<b$. Let $n>1$, and suppose that the mapping $g:[a, b] \longrightarrow \mathbb{R}^n$ is continuous and that $g$ is differentiable on the open interval $(a, b)$. It is tempting to generalize the mean value theorem (Theorem 1.2.3) to the assertion

$$
g(b)-g(a)=g^{\prime}(c)(b-a) \quad \text { for some } c \in(a, b) .
$$

The assertion is grammatically meaningful, since it posits an equality between two $n$-vectors. The assertion would lead to a slight streamlining of the proof of Lemma 5.1.3, since there would be no need to reduce to scalar output. However, the assertion is false.
(a) Let $g:[0,2 \pi] \longrightarrow \mathbb{R}^2$ be $g(t)=(\cos t, \sin t)$. Show that (5.1) fails for this $g$. Describe the situation geometrically.
(b) Let $g:[0,2 \pi] \longrightarrow \mathbb{R}^3$ be $g(t)=(\cos t, \sin t, t)$. Show that (5.1) fails for this $g$. Describe the situation geometrically.
(c) Here is an attempt to prove (5.1): Let $g=\left(g_1, \ldots, g_n\right)$. Since each $g_i$ is scalar-valued, we have for $i=1, \ldots, n$ by the mean value theorem,

$$
g_i(b)-g_i(a)=g_i^{\prime}(c)(b-a) \text { for some } c \in(a, b) .
$$

Assembling the scalar results gives the desired vector result.
What is the error here?

Daphne Pusey
Daphne Pusey
Numerade Educator