Continuity

Show your work for each problem.

1.
Which of these is NOT one of the three conditions for $f$ to be continuous at $x = a$ ?
1. a) $\lim_{x \to a} f(x)$ exists
2. b) $f(a)$ is defined
3. c) $\lim_{x \to a} f(x) = f(a)$
4. d) $f(a) = 0$
Answer: $f(a) = 0$
The three conditions are: $f(a)$ is defined, $\lim_{x \to a} f(x)$ exists, and the limit equals $f(a)$ . The value $f(a)$ can be any number; it does not have to be $0$ .
2.
$f(x) = 3x^2 - 2x + 5$ is continuous everywhere. Find $\lim_{x \to 2} f(x)$ .
Answer: 13
Because $f$ is continuous, $\lim_{x \to 2} f(x) = f(2) = 3 \cdot 4 - 4 + 5 = 13$ .
3.
Where is $f(x) = \frac{1}{x - 4}$ NOT continuous?
1. a) $x = 4$
2. b) $x = -4$
3. c) $x = 0$
4. d)It is continuous everywhere
Answer: $x = 4$
The denominator $x - 4$ is $0$ at $x = 4$ , so $f(4)$ is undefined there. Everywhere else $f$ is continuous.
4.
$g(x) = \frac{x^2 - 1}{x - 1}$ is undefined at $x = 1$ . Which value should $g(1)$ get so that $g$ is continuous at $x = 1$ ?
Answer: 2
$\frac{x^2 - 1}{x - 1} = \frac{(x - 1)(x + 1)}{x - 1} = x + 1$ for $x \neq 1$ , so the limit is $1 + 1 = 2$ . Setting $g(1) = 2$ makes the limit equal the value.
5.
Check whether $f(x) = x^2 + 2x$ is continuous at $x = 3$ .
Answer: $\lim_{x \to 3} f(x) = f(3) = 15$
1. What is $f(3)$ ? 15
2. What is $\lim_{x \to 3} f(x)$ ? 15
3. Is $f$ continuous at $x = 3$ ? yes
6.
Is $f(x) = \begin{cases} 2x + 1, & x < 1 \\ 4 - x, & x \geq 1 \end{cases}$ continuous at $x = 1$ ?
Answer: $\lim_{x \to 1} f(x) = f(1) = 3$
1. Left-hand limit $\lim_{x \to 1^-} f(x)$ ? 3
2. Right-hand limit $\lim_{x \to 1^+} f(x)$ ? 3
3. Value $f(1)$ ? 3
4. Is $f$ continuous at $x = 1$ ? yes
7.
Which function is continuous for ALL real numbers?
1. a) $\frac{1}{x}$
2. b) $e^x + x^3$
3. c) $\sqrt{x}$
4. d) $\ln x$
Answer: $e^x + x^3$
$\frac{1}{x}$ is undefined at $0$ , $\ln x$ only exists for $x > 0$ and $\sqrt{x}$ only for $x \geq 0$ . $e^x + x^3$ is a sum of two functions that are continuous everywhere, so it is continuous everywhere.
8.
Find $k$ so that $f(x) = \begin{cases} x + k, & x < 2 \\ x^2, & x \geq 2 \end{cases}$ is continuous at $x = 2$ .
Answer: 2
Left: $2 + k$ . Right and value: $2^2 = 4$ . So $2 + k = 4$ and $k = 2$ .
9.
Find $k$ so that $f(x) = \begin{cases} kx - 2, & x \leq 3 \\ x^2 + 1, & x > 3 \end{cases}$ is continuous everywhere.
Answer: 4
1. Right-hand limit at $x = 3$ (use $x^2 + 1$ )? 10
2. The left piece gives $3k - 2$ . From $3k - 2 = 10$ , what is $3k$ ? 12
3. What is $k$ ? 4
10.
Let $h(x) = \frac{x^2 - 5x + 6}{x - 2}$ for $x \neq 2$ and $h(2) = 3$ . Is $h$ continuous at $x = 2$ ?
Answer: $\lim_{x \to 2} h(x) = -1 \neq h(2) = 3$
1. What is $h(2)$ ? 3
2. After factoring and canceling, $h(x) = x - 3$ . What is $\lim_{x \to 2} h(x)$ ? -1
3. Is $h$ continuous at $x = 2$ ? no
11.
Where is $f(x) = \frac{x + 3}{x^2 - 9}$ continuous?
1. a)All real numbers except $x = -3$
2. b)All real numbers except $x = 3$ and $x = -3$
3. c)All real numbers
4. d)All real numbers except $x = 3$
Answer: All real numbers except $x = 3$ and $x = -3$
$x^2 - 9 = (x - 3)(x + 3)$ is $0$ at $x = 3$ and $x = -3$ , so $f$ is undefined at both. At $x = -3$ the graph has a hole, at $x = 3$ a vertical asymptote, but in both cases $f$ is not continuous there.
12.
At how many values of $x$ is $f(x) = \frac{x - 1}{x^3 - 4x}$ not continuous?
Answer: 3
$x^3 - 4x = x(x - 2)(x + 2)$ is $0$ at $x = 0$ , $x = 2$ and $x = -2$ . The numerator does not cancel any of them, so $f$ is discontinuous at exactly $3$ values.
13.
Use the Intermediate Value Theorem to decide whether $f(x) = x^3 - 2x - 5$ has a root in $[2, 3]$ .
Answer: $f(2) = -1 < 0 < 16 = f(3)$
1. What is $f(2)$ ? -1
2. What is $f(3)$ ? 16
3. Does the IVT guarantee a root in $[2, 3]$ ? yes
14.
Find $a$ so that $f(x) = \begin{cases} ax + 3, & x < 2 \\ x^2 + a, & x \geq 2 \end{cases}$ is continuous everywhere.
Answer: 1
Left: $2a + 3$ . Right and value: $4 + a$ . Then $2a + 3 = 4 + a$ , so $a = 1$ . Check: both sides give $5$ .
15.
Find $a$ and $b$ so that $f(x) = \begin{cases} x + 1, & x < 1 \\ ax + b, & 1 \leq x < 3 \\ x^2 - 3, & x \geq 3 \end{cases}$ is continuous everywhere.
Answer: a = 2, b = 0
1. At $x = 1$ the left piece gives $1 + 1$ . So $a + b = ?$ 2
2. At $x = 3$ the right piece gives $3^2 - 3$ . So $3a + b = ?$ 6
3. Subtract the first equation from the second. What is $a$ ? 2
4. What is $b$ ? 0
16.
$f(x) = \frac{1}{x}$ has $f(-1) = -1$ and $f(1) = 1$ , but $f(x) = 0$ has no solution. Why does the Intermediate Value Theorem not apply on $[-1, 1]$ ?
1. a) $f(-1)$ and $f(1)$ have the same sign
2. b)The theorem only works for polynomials
3. c) $f$ is not continuous on $[-1, 1]$ : it is undefined at $x = 0$
4. d) $0$ is not between $-1$ and $1$
Answer: $f$ is not continuous on $[-1, 1]$ : it is undefined at $x = 0$
The IVT needs $f$ to be continuous on the whole closed interval. $\frac{1}{x}$ is undefined at $x = 0$ , which lies inside $[-1, 1]$ , so the theorem says nothing here.
17.
Let $f(x) = \frac{\sqrt{x} - 2}{x - 4}$ for $x \neq 4$ and $f(4) = k$ . Find $k$ so that $f$ is continuous at $x = 4$ .
Answer: 1/4
1. Substitute $x = 4$ into the numerator $\sqrt{x} - 2$ . What do you get? 0
2. $(\sqrt{x} - 2)(\sqrt{x} + 2) = x - c$ . What is $c$ ? 4
3. So $f(x) = \frac{1}{\sqrt{x} + 2}$ . What is $\lim_{x \to 4} f(x)$ ? Write it as a fraction. 1/4
4. What must $k$ be? 1/4
18.
How many roots of $f(x) = x^3 - 3x + 1$ in $[0, 2]$ does the Intermediate Value Theorem guarantee, using the points $x = 0$ , $1$ , $2$ ?
Answer: 2
1. What is $f(0)$ ? 1
2. What is $f(1)$ ? -1
3. What is $f(2)$ ? 3
4. How many sign changes, and so how many guaranteed roots? 2