notes / updates
October 3, 2024 | Quiz Feedback
Below find some quiz notes. Graded work will be returned in class. If there are mistakes or typos below, feel free to email me.
Hi everyone. Today's quiz went well on average.
General comments:
Check where "things go" and use arrows to show it.
Use limit signs.
Use equal signs.
Do not use equal signs between the limit sign and the expression.
Specific issues:
The only problem that presented a challenge of sorts was \( \lim\limits_{x\to1} \frac{x^3-1}{x^2-1} \)
After checking that both numerator and denominator go to zero, use the difference of squares for the bottom and the difference of cubes formula for the top.
\( A^3 - B^3 = (A-B)(A^2 + AB + B^2) \)
\( \lim\limits_{x\to1} \frac{x^3-1}{x^2-1} = \lim\limits_{x\to1} \frac{(x-1)(x^2+x+1)}{(x-1)(x+1)} = \lim\limits_{x\to1} \frac{x^2+x+1}{x+1} = \frac{3}{2} \)
The current grade on myCushing is up to date as of October 3, 7:25pm. Graded quizzes will be handed back in class on Friday.
September 26, 2024 | Quiz Notes
Below find some quiz notes. Graded work will be returned in class. If there are mistakes or typos below, feel free to email me.
P1
No issues except make sure there is a hole at the point (2, 3), so the graph should approach y=3 from both sides. $f(2)$, however, equals 1, so we draw a solid point at (2, 1).
P2
Let's get in the habit of drawing neatly. Label your x- and y-axis with arrows. The problem also asked for which values of
a does \( \lim\limits_{x\to a} f(x) \) exist. the overall limit exists at all real numbers in the domain except at 0 and 1. At these two places, the one-sided and right-sided limits do not equal each other, so the overall limit does not exist.
September 23, 2024 | Quiz Notes
Below find some quiz notes. Graded work will be returned in class. If there are mistakes or typos below, feel free to email me.
P1
\(f(x)= 4x^3 -x^2-12x + 3 \)
a) \( f(x)=x^2 (4x-1) - 3 (4x-1) = (x^2-3)(4x-1) = (x-\sqrt{3})(x+\sqrt{3}) (4x-1) \)
b) x-intercepts: \( (\pm \sqrt{3}, 0), (1/4,0) \)
y-intercept: \( (0, 3) \)
c) Sketch:
d) The domain is all real numbers except for the x-intercepts of f(x), since the denominator cannot be zero.
\( (-\infty, -\sqrt{3}) \cup (-\sqrt{3}, 1/4) \cup (1/4, \sqrt{3}) \cup (\sqrt{3}, \infty) \)
P2
Factor completely: \( (\sec x - 2) (\sec x + 1) = 0 \)
Solve: \( \sec x = 2 \text{ or } \sec x = -1 \)
Find three solutions on the given domain: \( x= \pi/3, 5\pi/3, \pi \)
September 22, 2024 | Weekly updates & Current grades
Hi everyone. I write to clarify a couple of things regarding grades and grading.
First, on myCushing under your progress tab you should be able to see your current grade for this course, which is a weighted average of the assessments we have written so far. While individual quiz scores will not be published on myCushing, the current grade will be updated approximately once week.
Second, when you read your weekly update, be sure to check the comments as well as the grade. Sometimes I add individual feedback based on recent performances or in-class observations.
As always, let me know if you have any questions.
September 17, 2024 | Notes on recent quiz
If there are typos or mistakes, let me know so I can fix them here.
P1.
The domain of the function \( f(x) = \log(x^2-1) \) can be determined by setting \( x^2 - 1 > 0 \). This expression holds true whenever \( x < - 1 \text{ or } x > 1 \).
Using interval notation, the domain is \( (-\infty, -1) \cup (1, \infty) \).
P2.
This was similar to the problems we did in class. Combine into a single logarithm:
\( \log_{2}((x+2)(x-1)) = 2 \)
Solve the quadratic equation to find : \( x=-3, x = 2 \). Of these, only \( x=2 \) is in the domain.
P3.
\( f(x) = \log_3(3-x) \)
domain: \( x<3 \)
range: all reals
intercepts: (0,1) and (2, 0)
vertical asymptote: \( x=3 \)
Graph:
September 16, 2024 | Notes on recent quiz
If there are typos or mistakes, let me know so I can fix them here.
P1.
a) \( f(x) = \sqrt{(1-x)(x+3)} \rightarrow (1-x)(x+3) \geq 0 \)
Solve graphically or algebraically or on the number line with test points, to conclude that the domain is: \( [-3, 1] \)
b) \( g(x) = \sqrt{-x-2} \rightarrow -x-2 \geq 0 \). The domain is: \( (-\infty, -2] \)
c) The domain of \( h(x) = f(x) + g(x) \) is the intersection of the two domains above, which is \( [-3, -2] \).
P2.
a) The function \( f(x) = \frac{x}{1-x^2} \) is odd because: \( f(-x) = \frac{-x}{1-(-x)^2} = - \frac{x}{1-x^2} = -f(x) \).
Note that you could also test specific values here but the steps above are more convincing.
b) Similar to the first problem, we can form a function such as: \( f(x) = \sqrt{(1-x)(x-5)}\). There are other examples that work, too.
September 10, 2024
At the following
link (opens new window) find an online precalculus textbook. The content in chapters 1 through 7 covers a great deal of content we need to know well in order to be successful in this course.
September 9, 2024
AP registration instructions at this [
link ].
For your course section, use join code: MY274M.
If you have any questions, contact Mr. Dan Monahan.
September 1, 2024
Hi everyone. In this area of the course page we will post notes, solutions, reminders, and links to other resources throughout the year. Entries will be dated with the most recent posted at the top of the page.