MATH 409 Lecture 21

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End Exam 2 content

Lecture Slides

Exam 2 Review

Topics

Derivatives

• Derivative of a function
• Differentiability Theorems
• derivative of inverse function
• Mean Value Theorem
  • Rolle's Theorem
  • Generalized Mean Value Theorem
• Taylor's Formula
• l'Hôpital's rule

Integrals

• Definitions
  • Darboux Sums
  • Riemann Sums
  • Riemann Integral
• Properties of Integrals
• Fundamental Theorem of Calculus (both parts)
• Integration by Parts
• Change of variable in an integral

Chapters 4.1–4.5, 5.1–5.3

Differentiability Theorems

• sum / difference rules
• product rule
• quotient rule
• chain rule
• differentiability implies continuity
• Rolle's Theorem: If a function $f$ is continuous on a closed interval $[a,b]$, differentiable on $(a,b)$, and $f(a)=f(b)$, then $f'(c)=0$ for some $c\in (a,b)$
• Mean Value Theorem: If a function $f$ is continuous on $[a,b]$ and differentiable on $(a,b)$, then there exists $c\in (a,b)$ such than $f(b)-f(a)=f'(c)\,(b-a)$.
• $f$ is increasing on $[a,b]$ if and only if $f'\geq 0$ on $(a,b)$
• $f$ is decreasing on $[a,b]$ if and only if $f'\leq 0$ on $(a,b)$
• $f$ is constant on $[a,b]$ if and only if $f'=0$ on $(a,b)$

Properties of Integrals

• Linearity:
  • $\int _{a}^{b}\left(f(x)+g(x)\right)\,\mathrm {d} x=\int _{a}^{b}f(x)\,\mathrm {d} x+\int _{a}^{b}g(x)\,\mathrm {d} x$
  • $\int _{a}^{b}\left(\alpha \,f(x)\right)\,\mathrm {d} x=\alpha \,\int _{a}^{b}f(x)\,\mathrm {d} x$
• Subinterval property: $\int _{a}^{b}f(x)\,\mathrm {d} x=\int _{a}^{c}f(x)\,\mathrm {d} x+\int _{c}^{b}f(x)\,\mathrm {d} x$
• Comparison Theorem: If $f(x)\leq g(x)$ for all $x\in [a,b]$, then $\int _{a}^{b}f(x)\,\mathrm {d} x\leq \int _{a}^{b}g(x)\,\mathrm {d} x$

Fundamental Theorem of Calculus

• Part 1: $F(x)=\int _{a}^{x}f(t)\,\mathrm {d} t$, $x\in [a,b]$ is continuously differentiable on $[a,b]$ and $F'(x)=f(x)$ for all $x\in [a,b]$
• Part 2: If a function $F$ is differentiable on $[a,b]$, and the derivative $F'$ is integrable on $[a,b]$, then $\int _{a}^{x}F'(t)\,\mathrm {d} t=F(x)-F(a)$ for all $x\in [a,b]$.

Sample Problems

Problem 1: Prove the Chain Rule

Theorem. If a function $f$...

Proved in class (except for one small part)

Problem 2: Find the Limits

$\lim _{x\to 0}\left(1+x\right)^{\frac {1}{x}}$

The function $f(x)=\left(1+x\right)^{\frac {1}{x}}$ is well-defined on $(-1,\infty )$ except at $0$. Since $f(x)>0$ for all $x>1$, a function $g(x)=\log {f(x)}$ is well-defined on $(-1,\infty )$ except at $0$ as well. For any $x>1$, we have $g(x)=\log {\left(1+x\right)^{\frac {1}{x}}}=x^{-1}\,\log {\left(1+x\right)}$. Hence $g={\frac {h_{1}}{h_{2}}}$, where $h_{1}(x)=\log {\left(1+x\right)}$ and Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle h_2(x) = x} are continuously differentiable on Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \left( -1, \infty \right)} . Since $h_{1}(0)=h_{2}(0)=0$, it follows that Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \lim_{x \to 0} h_1(x) = \lim_{x \to 0} h_2(x) = 0} . By l'Hôpital's rule, $h_{1}'(0)=1$ and Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle h_2'(0) = 1} , so Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \lim_{x \to 0} g(x) = \lim_{x \to 0} \frac{h_1(x)}{h_2(x)} = \frac{1}{1} = 1}

Since $f=\mathrm {e} ^{g(x)}$, a composition of Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle g} with a continuous function, it follows that Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \lim_{x \to 0} f(x) = \mathrm{e}^{1} = \mathrm{e}}

$\lim _{x\to +\infty }\left(1+x\right)^{\frac {1}{x}}$ Similar to above, Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \lim_{x \to +\infty} h_1(x) = \lim_{x \to +\infty} h_2(x) = +\infty} . At the same time, $h_{1}'(x)\to 0$ as Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle x \to +\infty} , while Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle h_2'} is identically Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle 1} . Using l'Hôpital's Rule, we obtain $\lim _{x\to +\infty }g(x)=\lim _{x\to +\infty }{\frac {h_{1}(x)}{h_{2}(x)}}={\frac {0}{1}}=0$

Since $f=\mathrm {e} ^{g(x)}$, a composition of Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle g} with a continuous function, it follows that Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \lim_{x \to 0} f(x) = \mathrm{e}^{0} = 1}

Problem 3: Limit of a sequence

Find the limit of a sequence Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle x_n = \frac{1^k + 2^k + \dots + n^k}{n^{k+1}}} for Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle n \in \mathbb{N}} , where $k$ is a natural number.

The general element of the sequence can be represented as

Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle x_n = \frac{1^k + 2^k + \dots + n^k}{n^k} \cdot \frac{1}{n} = \left( \frac{1}{n} \right)^k \frac{1}{n} + \left( \frac{2}{n} \right)^k \frac{1}{n} + \dots + \left( \frac{n}{n} \right)^k \frac{1}{n}}

which shows that $x_{n}$ is a Riemann sum of the function Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle f(x) = x^k} on the interval Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle [0,1]} that corresponds to the partition $P_{n}=\left\{0,{\frac {1}{n}},{\frac {2}{n}},\ldots ,{\frac {n-1}{n}},1\right\}$ and samples Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle t_j = \frac{j}{n}} , Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle j = 1, 2, \ldots, n} . The norm of the partition is Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \left\| P_n \right\| = \frac{1}{n}} . Since Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \left\| P_n \right\| \to 0} as $n\to \infty$ and the function Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle f} is integrable on Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle [0,1]} , the Riemann sums Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle x_n} converge to the integral:

Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \lim_{n \to \infty} x_n = \int_{0}^{1} x^k \,\mathrm{d}x = \left. \frac{x^{k+1}}{k+1} \right|_{x=0}^1 = \frac{1}{k+1}}

Problem 4: Find/evaluate indefinite/definite integrals

Subproblem 1

$\int {\frac {x^{2}}{1-x}}\,\mathrm {d} x$

A standard way to evaluate this type of function is to split it into the sum of a polynomial and a simple fraction:

${\frac {x^{2}}{1-x}}={\frac {x^{2}-1+1}{1-x}}={\frac {x^{2}-1}{1-x}}+{\frac {1}{1-x}}=-x-1-{\frac {1}{x-1}}$

Since the domain of the function is $(-\infty ,1)\cup (1\infty )$, the indefinite integral has differetn representations on the intervals $(-\infty ,1)$ and $(1,\infty )$:

$\int {\frac {x^{2}}{1-x}}={\begin{cases}-{\frac {x^{2}}{2}}-x-\log {\left(1-x\right)}+C_{1}&x<1\\-{\frac {x^{2}}{2}}-x-\log {\left(x-1\right)}+C_{2}&x>1\end{cases}}$

Subproblem 2

$\int _{0}^{\pi }\sin ^{2}{(2x)}\,\mathrm {d} x$

To integrate this function, we use a trigonometric formula $1-\cos {2\alpha }=2\sin ^{2}{\alpha }$ and a new variable $u=4x$:

$\int _{0}^{\pi }\sin ^{2}{(2x)}\,\mathrm {d} x=\int _{0}^{\pi }{\frac {1-\cos {(4x)}}{2}}\,\mathrm {d} x=\int _{0}^{\pi }{\frac {1-\cos {(4x)}}{8}}\,\mathrm {d} (4x)=\int _{0}^{4\pi }{\frac {1-\cos {u}}{8}}\,\mathrm {d} u=\left.{\frac {u-\sin {u}}{8}}\right|_{u=0}^{4\pi }={\frac {\pi }{2}}$

Subproblem 3

$\int \log ^{3}{x}\,\mathrm {d} x$

To find this indefinite integral, we integrate by parts:

${\begin{aligned}\int \log ^{3}{x}\,\mathrm {d} x&=x\,\log ^{3}{x}-\int x\,\mathrm {d} (\log ^{3}{x})\\&=x\,\log ^{3}{x}-\int x\left(\log ^{3}{x}\right)'\,\mathrm {d} x\\&=x\,\log ^{3}{x}-\int 3\log ^{2}{x}\,\mathrm {d} x\\&=x\,\log ^{3}{x}-3x\,\log ^{2}{x}+\int x\,\mathrm {d} \left(3\log ^{2}{x}\right)\\&=x\,\log ^{3}{x}-3x\,\log ^{2}{x}+\int 6\log {x}\,\mathrm {d} x\\&=x\,\log ^{3}{x}-3x\,\log ^{2}{x}+6x\,\log {x}-\int x\,\mathrm {d} \left(6\log {x}\right)\\&=x\,\log ^{3}{x}-3x\,\log ^{2}{x}+6x\,\log {x}-\int 6\,\mathrm {d} x\\&=x\,\log ^{3}{x}-3x\,\log ^{2}{x}+6x\,\log {x}-6x+C\end{aligned}}$

Subproblem 4

$\int _{0}^{\frac {1}{2}}{\frac {x}{\sqrt {1-x^{2}}}}\,\mathrm {d} x$

To integrate this function, we introduce a new variable $u=1-x^{2}$:

${\begin{aligned}\int _{0}^{\frac {1}{2}}{\frac {x}{\sqrt {1-x^{2}}}}\,\mathrm {d} x&=-{\frac {1}{2}}\,\int _{0}^{\frac {1}{2}}{\frac {\left(1-x^{2}\right)'}{\sqrt {1-x^{2}}}}\,\mathrm {d} x\\&=-{\frac {1}{2}}\,\int _{0}^{\frac {1}{2}}{\frac {1}{\sqrt {1-x^{2}}}}\,\mathrm {d} \left(1-x^{2}\right)\\&=-{\frac {1}{2}}\,\int _{1}^{\frac {3}{4}}{\frac {1}{\sqrt {u}}}\,\mathrm {d} u\\&=\ldots \\&=1-{\frac {\sqrt {3}}{2}}\end{aligned}}$

Subproblem 5

$\int _{0}^{1}{\frac {1}{\sqrt {4-x^{2}}}}\,\mathrm {d} x$

To integrate this function, we use a substitution $x=2\sin {t}$. Observe that $x$ changes from Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle 0} to Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle 1} when Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle t} changes from $0$ to ${\frac {\pi }{6}}$):

Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \begin{align} \int_{0}^{1} \frac{1}{\sqrt{4-x^2}} \,\mathrm{d}x &= \int_{0}^{\frac{\pi}{6}} \frac{1}{\sqrt{4-(2\sin{t})^2}} \, \mathrm{d}(2\sin{t}) \\ &= \int_{0}^{\frac{\pi}{6}} \frac{\left( 2\sin{t} \right)}{\sqrt{4-4\sin^2{t}}} \,\mathrm{d}t \\ &= \int_{0}^{\frac{\pi}{6}} \frac{2\cos{t}}{\sqrt{4\cos^2{t}}} \,\mathrm{d}t \\ &= \int_{0}^{\frac{\pi}{6}} 1 \,\mathrm{d}x \\ &= \frac{\pi}{6} \end{align}}

Bonus Problem 5

Suppose Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle p : \mathbb{R} \to \mathbb{R}} is locally a polynomial, which means that for every Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle c \in \mathbb{R}} , there exists Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \epsilon > 0} such that $p$ coincides with a polynomial on the interval $\left(c-\epsilon ,c+\epsilon \right)$. Prove that $p$ is a polynomial.

Proof. For any Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle c \in \mathbb{R}} let Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle p_c} denote a polynomial and Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \epsilon_c} denote a positive number such that Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle p(x) = p_c(x)} for all $x\in (c-\epsilon _{c},c+\epsilon _{c})$. Consider two sets:

Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \begin{align} E_+ &= \left\{ x > 0 ~\mid~ p(x) \ne p_0(x) \right\} \\ E_- &= \left\{ x < 0 ~\mid~ p(x) \ne p_0(x) \right\} \end{align}}

We are going to show that Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle E_+ = E_- = \emptyset} .

Assume that the set $E_{+}$ is not empty. Clearly, Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle E_+} is bounded below, hence Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle d = \inf{E_+}} is a well-defined real number. Note that $E_{+}\subset \left[\epsilon _{0},\infty \right)$ Therefore Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle d \ge \epsilon_0 > 0} .

Observe that $p(x)=p_{0}(x)$ for Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle x \in (0,d)} and Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle p(x) = p_d(x)} for Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle x \in \left( d - \epsilon_d, d + \epsilon_d \right)} . The interval $(0,d)$ overlaps with the interval $\left(d-\epsilon _{d},d+\epsilon _{d}\right)$. Hence Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle p_d} coincides with Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle p_0} on the intersection Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle (0,d) \cap (d - \epsilon_d, d + \epsilon_d)} . Equivalently, the difference Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle p_d - p_0} is zero on $(0,d)\cap (d-\epsilon _{d},d+\epsilon _{d})$. Since $p_{d}-p_{0}$ is a polynomial and any nonzero polynomial has only finitely many roots, we conclude that Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle p_d - p_0} is identically 0. Then the polynomials $p_{d}$ and Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle p_0} are the same. It follows that Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle p(x) = p_0(x)} for Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle x \in (0,d+\epsilon_d)} , so $d\neq \inf {E_{+}}$, a contradiction.

Thus $E_{+}=\emptyset$. Similarly, we prove that the set Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle E_-} is empty as well. Since $E_{+}=E_{-}=\emptyset$, the function Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle p} coincides with the polynomial Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle p_0} everywhere.

Bonus Problem 6

Show that a function Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle f(x) = \begin{cases} \mathrm{e}^{-\frac{1}{1-x^2}} & \left| x \right| < 1 \\ 0 & \left| x \right| \ge 1 \end{cases}}

is infinitely differentiable on Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \mathbb{R}} .

Observe that the integral of this function goes from 0 to 1 smoothly. Constant functions and Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \mathrm{e}^{f(x)}} are infinitely differentiable.