Show each step of your work and fully simplify each expression.
Turn in your answers in class on a physical piece of paper.
Staple multiple sheets together.
Feel free to use Desmos for graphing.
Answer the following:
Read (or watch the recorded lecture) these two lecture notes on trigonometric functions: 5.1: Unit Circle and 5.2: Trigonometric Functions. This is mandatory reading if you do not remember trigonometric functions. Draw a checkmark to get credit for this problem.
Using the unit circle, explain why $\tan\left(\dfrac{\pi}{2}\right)$ is undefined.
Find the six trigonometric functions of $t = \dfrac{2\pi}{3}$.
Find the six trigonometric functions of $t = -\dfrac{5\pi}{6}$.
Suppose \[f(x) = \sin(x) + \cos(x) \qquad g(x) = \tan(x)\]
Evaluate and simplify the following expressions without a calculator/the internet:
You are given a function $F(x)$. Decompose $F(x)$ into two functions $f, g$ where $F = f \circ g$.
$F(x) = \sin(\cos x)$
$F(x) = \sin^2(x)$
$F(x) = \sin(x^2)$
$F(x) = (x^3 - x^2 - 1)^{2/3}$
$F(x) = (x^2 - x)^2$
$F(x) = \sqrt[5]{(x + 1)^3}$
$F(x) = \sec(\tan(x))$
Find the domain of the following functions:
$f(x) = \dfrac{1}{x^2 - 1}$
$f(x) = \sqrt{x} + \dfrac{1}{x}$
$f(x) = \sin(x)$
$f(x) = \dfrac{1}{\sqrt{x}}$
Hint: Both problems are present.
Suppose \[f(x) = \sin(x) \qquad g(x) = x^2 - x \qquad h(x) = f(x)g(x)\] Evaluate the following:
$f\circ g$
$g \circ f$
$h \circ f$
Suppose $f(x)$ is a function. Explain in English the geometric intuition behind the symbols \[\lim_{x\rightarrow a}f(x) = L\] means.
Suppose \[\lim_{x\rightarrow a}f(x) = L\] Does the limit see what happens exactly at the $x$-value $a$?
Suppose we have an expression $\dfrac{1}{x}$ where $x$ gets closer and closer to 0 from the right. What does the expression $\dfrac{1}{x}$ end up approaching?
A function $f(x)$ has the following graph:
Find the following. If they do not exist explain why.
$\displaystyle\lim_{x\rightarrow 2^-} f(x)$
$\displaystyle\lim_{x\rightarrow 2^+} f(x)$
$\displaystyle\lim_{x\rightarrow 2} f(x)$
$f(2)$
$\displaystyle\lim_{x\rightarrow 4} f(x)$
$f(4)$
Sketch one graph of a function $f(x)$ which satisfies the following:
$f(0) = -1$
$f(3) = 1$
$\displaystyle\lim_{x\rightarrow 0} f(x) = 1$
$\displaystyle\lim_{x\rightarrow 3^-} f(x) = -2$
$\displaystyle\lim_{x\rightarrow 3^+} f(x) = 2$
A function $g(x)$ has the following graph:
Find the following. If they do not exist explain why.
$\displaystyle\lim_{x\rightarrow -3} g(x)$
$\displaystyle\lim_{x\rightarrow 2^-} g(x)$
$\displaystyle\lim_{x\rightarrow 2^+} g(x)$
$\displaystyle\lim_{x\rightarrow 2} g(x)$
$\displaystyle\lim_{x\rightarrow -1} g(x)$
$g(3)$
Given the following functions $f(x)$ and $g(x)$:
Find the following limits, if it exists. If it doesn't, explain why.
If a function $f(x)$ is continuous at $x = a$, what does $\displaystyle \lim_{x\rightarrow a} f(x)$ have to be?
Suppose a function $f(x)$ has $f(2) = 5$ and $\displaystyle \lim_{x \rightarrow 2}f(x) = 5$. Is the function continuous at $x = 2$?
In the context of finding a limit, why does continuity make the problem much easier?
Use the three-part definition of continuity to explain why the function is continuous at the given number $a$.
$\displaystyle f(x) = \begin{cases}(x-1)^2& x < 0 \\ (x+1)^2 & x \geq 0\end{cases} \qquad a = 0$
$\displaystyle f(x) = \begin{cases}x(x-1)& x < 1 \\ 0 & x = 1 \\ \sqrt{x-1} & x > 1\end{cases} \qquad a = 1$
Use the three-part definition of continuity to explain why the function is discontinuous at the given number $a$. Sketch the graph of the function.
$\displaystyle f(x) = \begin{cases}x + 1 & x < 0 \\ x^2 & x \geq 0\end{cases} \qquad a = 0$
$\displaystyle f(x) = \begin{cases}x + 5 & x < 0 \\ 2 & x = 0 \\ -x^2 + 5 & x > 0\end{cases} \qquad a = 0$
$\displaystyle f(x) = \begin{cases}-x & x < 0 \\ 1 & x = 0 \\ x & x > 0\end{cases} \qquad a = 0$
How would you define $f(2)$ in the function \[f(x) = \dfrac{x^2 - x - 2}{x - 2}\] in order to make $f(x)$ continuous at $x = 2$?
Where is the function \[f(x) = \dfrac{\sin(x)}{(x^2 - 1)(x^2 - 4)}\] continuous?
What's the difference between a removeable, jump and infinite discontinuity?
Draw one graph of a function which satisfies the following conditions simultaneously:
Jump discontinuity at $2$ but continuous from the right at $2$
Discontinuous at $-1$ and $4$ but continuous from the left at $-1$ and from the right at $4$.
Continuous everywhere else
Suppose you are asked to find the limit \[\lim_{x\rightarrow \pi} \dfrac{\sin^2(x)\cos^3(x)}{x^{52} - 4x^{49} + x^{32} - x^3 + x^2 + 44}\]
You know the function is continuous at $x = \pi$. Find the limit.