The problem consists of defining concepts related to functions, determining the truth of statements about continuity and differentiability, and verifying a limit using the epsilon-delta definition and finding values for constants in a piecewise function to ensure continuity. Question A1 asks for definitions: a) $c \in D_f$ is a critical number of a function $f$. b) $\lim_{x \to c^+} f(x) = L$, where $L \in \mathbb{R}$ and $c \in \mathbb{R}$ is an accumulation point of $D_f$. c) A function $f$ has a local minimum at $c \in D_f$. Question A2 asks to determine if these statements are true or false: a) If $f$ is continuous at $c \in I$, then $f$ is differentiable at $c$. b) If both $\lim_{x \to c^-} f(x)$ and $\lim_{x \to c^+} f(x)$ exist, then $f$ is continuous at $c$. Question A3 has two parts: a) Verify $\lim_{x \to 2} (x^2 - 7) = -3$ using the $\epsilon$-$\delta$ definition. b) Find the values of $a, b \in \mathbb{R}$ that make $f(x)$ continuous on its domain, where $f(x)$ is defined as: $f(x) = \begin{cases} \frac{x^2 - 9}{x - 3} & \text{if } x < 3 \\ ax^2 + bx + 5 & \text{if } 3 \le x < 5 \\ 3x + a - b & \text{if } x \ge 5 \end{cases}$

1. Problem Description

The problem consists of defining concepts related to functions, determining the truth of statements about continuity and differentiability, and verifying a limit using the epsilon-delta definition and finding values for constants in a piecewise function to ensure continuity.

Question A1 asks for definitions:

a) $c \in D_f$ is a critical number of a function $f$.

b) $\lim_{x \to c^+} f(x) = L$, where $L \in \mathbb{R}$ and $c \in \mathbb{R}$ is an accumulation point of $D_f$.

c) A function $f$ has a local minimum at $c \in D_f$.

Question A2 asks to determine if these statements are true or false:

a) If $f$ is continuous at $c \in I$, then $f$ is differentiable at $c$.

b) If both $\lim_{x \to c^-} f(x)$ and $\lim_{x \to c^+} f(x)$ exist, then $f$ is continuous at $c$.

Question A3 has two parts:

a) Verify $\lim_{x \to 2} (x^2 - 7) = -3$ using the $\epsilon$-$\delta$ definition.

b) Find the values of $a, b \in \mathbb{R}$ that make $f(x)$ continuous on its domain, where $f(x)$ is defined as:

$f(x) = \begin{cases}

\frac{x^2 - 9}{x - 3} & \text{if } x < 3 \\

ax^2 + bx + 5 & \text{if } 3 \le x < 5 \\

3x + a - b & \text{if } x \ge 5

\end{cases}$

2. Solution Steps

Question A1:

a) $c \in D_f$ is a critical number of a function $f$ if $f'(c) = 0$ or $f'(c)$ does not exist.

b) $\lim_{x \to c^+} f(x) = L$ if for every $\epsilon > 0$, there exists a $\delta > 0$ such that if $c < x < c + \delta$, then $|f(x) - L| < \epsilon$. Also, $c$ must be an accumulation point of the domain $D_f$ meaning every open interval containing $c$ contains a point other than $c$ in $D_f$.

c) A function $f$ has a local minimum at $c \in D_f$ if there exists an open interval $(a, b)$ containing $c$ such that $f(c) \le f(x)$ for all $x \in (a, b) \cap D_f$.

Question A2:

a) False. A counterexample is $f(x) = |x|$ at $x = 0$. $f(x)$ is continuous at $x = 0$, but $f'(0)$ does not exist.

b) False. The limits must not only exist but also be equal to the value of the function at the point for the function to be continuous. A function can have left and right limits exist but not equal to $f(c)$.

Question A3:

a) To verify $\lim_{x \to 2} (x^2 - 7) = -3$, we need to show that for any $\epsilon > 0$, there exists a $\delta > 0$ such that if $0 < |x - 2| < \delta$, then $|(x^2 - 7) - (-3)| < \epsilon$.

We have $|(x^2 - 7) - (-3)| = |x^2 - 4| = |(x - 2)(x + 2)| = |x - 2||x + 2|$.

We want $|x - 2||x + 2| < \epsilon$.

Assume $|x - 2| < 1$. Then $-1 < x - 2 < 1$, so $1 < x < 3$, and $3 < x + 2 < 5$. Thus, $|x + 2| < 5$.

So $|x - 2||x + 2| < 5|x - 2|$.

If we choose $\delta = \min(1, \frac{\epsilon}{5})$, then if $|x - 2| < \delta$, we have $|(x^2 - 7) - (-3)| = |x - 2||x + 2| < 5|x - 2| < 5(\frac{\epsilon}{5}) = \epsilon$.

b) For $f(x)$ to be continuous, we need to ensure continuity at $x = 3$ and $x = 5$.

At $x = 3$:

$\lim_{x \to 3^-} f(x) = \lim_{x \to 3^-} \frac{x^2 - 9}{x - 3} = \lim_{x \to 3^-} \frac{(x - 3)(x + 3)}{x - 3} = \lim_{x \to 3^-} (x + 3) = 6$.

$f(3) = a(3^2) + b(3) + 5 = 9a + 3b + 5$.

For continuity at $x = 3$, we require $9a + 3b + 5 = 6$, so $9a + 3b = 1$.

At $x = 5$:

$\lim_{x \to 5^-} f(x) = a(5^2) + b(5) + 5 = 25a + 5b + 5$.

$f(5) = 3(5) + a - b = 15 + a - b$.

For continuity at $x = 5$, we require $25a + 5b + 5 = 15 + a - b$, so $24a + 6b = 10$, or $12a + 3b = 5$.

We have the system of equations:

$9a + 3b = 1$

$12a + 3b = 5$

Subtracting the first equation from the second gives $3a = 4$, so $a = \frac{4}{3}$.

Substituting into the first equation gives $9(\frac{4}{3}) + 3b = 1$, so $12 + 3b = 1$, which means $3b = -11$, so $b = -\frac{11}{3}$.

3. Final Answer

Question A1:

a) $c \in D_f$ is a critical number of a function $f$ if $f'(c) = 0$ or $f'(c)$ does not exist.

b) $\lim_{x \to c^+} f(x) = L$ if for every $\epsilon > 0$, there exists a $\delta > 0$ such that if $c < x < c + \delta$, then $|f(x) - L| < \epsilon$. Also, $c$ must be an accumulation point of the domain $D_f$ meaning every open interval containing $c$ contains a point other than $c$ in $D_f$.

c) A function $f$ has a local minimum at $c \in D_f$ if there exists an open interval $(a, b)$ containing $c$ such that $f(c) \le f(x)$ for all $x \in (a, b) \cap D_f$.

Question A2:

a) False.

b) False.

Question A3:

a) Verified by choosing $\delta = \min(1, \frac{\epsilon}{5})$.

b) $a = \frac{4}{3}$, $b = -\frac{11}{3}$.