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:
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.
$\displaystyle f(x) = \begin{cases}x + 1 & x < 0 \\ x^2 & x \geq 0\end{cases} \qquad a = 0$
$\displaystyle f(x) = \begin{cases}x + 5 & x < 0 \\ 2 & x = 0 \\ -x^2 + 5 & x > 0\end{cases} \qquad a = 0$
$\displaystyle f(x) = \begin{cases}-x & x < 0 \\ 1 & x = 0 \\ x & x > 0\end{cases} \qquad a = 0$
If a function $f(x)$ is continuous at $x = a$, what does $\displaystyle \lim_{x\rightarrow a} f(x)$ have to be?
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 \[f(x) = \dfrac{\sin(x)}{x + 1}\] 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
If a function $f(x)$ is continuous at $x = a$, what does $\displaystyle \lim_{x\rightarrow a} f(x)$ have to be?
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 $x = 2$?
Hint: find the hole in $f(x)$ and "fill in" the hole.
State in interval notation where each of the following functions are continuous.
$f(x) = \dfrac{\sin(x)}{x^2}$
$f(x) = 4x^{32} - 8x^2 + x - 1$
$f(x) = x^{32894983} - 2x + \dfrac{1}{x}$
$f(x) = \dfrac{\sin^4(x)\cos^3(x)}{x^2 - 4}$
Use the continuity limit swap theorem to prove the following function is continuous at 4.
\[\displaystyle f(x) = \begin{cases} \sqrt{x} & x < 4 \\ 2\cos\left(x-4\right) & x \geq 4\end{cases} \qquad a = 4\]
Draw three different graphs, each of which shows one of each of 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
Show $f(x) = x^4 + x - 3$ has a root between $(1, 2)$.
Suppose $(1, f(1))$ is a point on the graph of $f(x)$. What is the equation of the tangent line at $x = 1$?
When we are using the definition of the slope of the tangent line at the point $(a, f(a))$:
What type of limit is this called?
What always happens to the $h$ in the denominator?
Suppose the above phenomena does not happen. What do you think went wrong?
Using the limit definition of the slope of a tangent line, i.e. \[m = \lim_{h\rightarrow 0} \dfrac{f(a + h) - f(a)}{h}\], find an equation of the tangent line for the following functions at the given point: