Notes
Definition
- $f$ is continuous at $a$ if $f(a)$ is defined, $\displaystyle \lim_{x \to a} f(x)$ exists, and the limit equals $f(a)$.
- Polynomials are continuous everywhere. Rational functions are continuous except where the denominator is 0.
Types of discontinuities
- Removable: a hole. The limit exists but the function is undefined or has a different value there.
- Jump: the one-sided limits exist but are different.
- Infinite: a vertical asymptote.
- To make a piecewise function continuous, set the pieces equal at the boundary and solve for the constant.
Intermediate Value Theorem
- If $f$ is continuous on $[a, b]$, it takes every value between $f(a)$ and $f(b)$.
- A sign change of a continuous function on $[a, b]$ guarantees a root in $(a, b)$.
Topics in this unit
Each topic has its own short notes and practice.
Practice
It simplifies to $x + 5$ with a hole at $x = 5$.
The function grows without bound near 0.
Set $3k + 1 = 10$, so $k = 3$.
The sign change and continuity guarantee at least one root. Nothing else is guaranteed.
The denominator is 0 at $x = -2$ and $x = 3$.
The others are undefined at 0. The polynomial is continuous everywhere.
Free response
Write your answer first, then open the scoring guide.
Free-response question
Let $f(x) = 2x + 3$ for $x < 1$ and $f(x) = x^2 + k$ for $x \geq 1$. Let $g(x) = x^3 - 3x + 1$.
(a) Find $\lim_{x \to 1^-} f(x)$.
Show the answer and scoring guide
Use the left piece: $2(1) + 3 = 5$.
Check off the points you earned:
(b) Find $k$ so that $f$ is continuous at $x = 1$.
Show the answer and scoring guide
Set $1^2 + k = 5$, so $k = 4$.
Check off the points you earned:
(c) Show that $g(x) = 0$ has a solution in $(1, 2)$.
Show the answer and scoring guide
$g$ is a polynomial, so it is continuous. $g(1) = 1 - 3 + 1 = -1$ and $g(2) = 8 - 6 + 1 = 3$. Since 0 is between $-1$ and 3, the IVT gives a $c$ in $(1, 2)$ with $g(c) = 0$.
Check off the points you earned: