notes / updates
October 5, 2024 | Quiz notes
P1
\( f(x)=\frac{3}{1-x} \text{ at } x=-2 \)
Since the problem asks specifically to use the definition of slope as a number, be sure to either
use the "h" definition: \( m = \lim\limits_{h\to 0} \frac{f(-2+h)-f(-2)}{h} \)
or the "a" definition: \( \lim\limits_{x\to -2} \frac{f(x)-f(-2)}{x-(-2)}\).
Do NOT use the derivative as a function definition: \( f'(x) = \lim\limits_{h\to 0} \frac{f(x+h)-f(x)}{h} \).
Be sure to show work for credit.
\( \lim_{x\to -2} \frac{f(x)-f(-2)}{x - (-2)} = \lim_{x\to -2} \frac{\frac{3}{1-x} - \frac{3}{3}}{x+2} = \lim_{x\to -2}\frac{6+3x}{3(1-x)(x+2)} = ... = \frac{1}{3} \)
Use the point-slope or slope-intercept form to find the equation of the tangent line: \( y = \frac{x+5}{3} \)
October 1, 2024 | Quiz feedback
a) Be sure to use limits when discussing continuity:
At \( x = 2 \) the graph of \(g \) is continuous from both sides, since \( \lim\limits_{x\to 2^-} g(x) = \lim\limits_{x\to 2^+} g(x) = 0 = g(2) \). Therefore the graph of \( g(x) \) is continuous at \(x = 2\).
At \( x = 3 \) the graph of \(g \) is continuous from both sides, since \( \lim\limits_{x\to 3^-} g(x) = \lim\limits_{x\to 3^+} g(x) = -1 = g(3) \). Therefore the graph of \( g(x) \) is continuous at \(x = 3 \).
At \( x = 4 \) the graph of \(g \) is only continuous from the right , since \( \lim\limits_{x\to 4^+} g(x) = \pi = g(4) \). \( g(x) \) is not continuous from the left at \(x = 4 \) because \( \lim\limits_{x\to 4^-} g(x) = 0 \neq g(4) = \pi \).
September 30, 2024 | Quiz notes
P1
Be sure to use arrows or explain your thinking in words. Be careful about plugging in right away!
a)
\( \lim\limits_{t\to \pi/2^+} e^{\tan t} = e^{ \lim\limits_{t\to \pi/2^+} \tan t} = 0 \) because \( \tan(t) \rightarrow - \infty \) and raising e to an infinitely small number tends towards zero.
b)
\( \lim\limits_{\theta \to 0}\tan(\frac{1-\cos\theta}{\theta}) = \tan(\lim\limits_{\theta \to 0}\frac{1-\cos\theta}{\theta} ) = \tan(0) = 0 \)
Note that the inner limit is that famous limit we talked about in class.
c)
Use the log properties here instead of subtracting infinities.
\(\lim\limits_{\theta\to 0^+} \ln(\sin(2\theta)) - \ln(\tan(\theta)) =\lim\limits_{\theta\to 0^+} \ln( \frac{ \sin(2\theta)}{\tan(\theta)}) = ... = \ln 2 \)
P2.
Let \( f(x) = x+\sin x + -1 \). This function is continuous on the interval \( [0, \pi/6] \), so we can use the Intermediate Value Theorem:
\( f(0) = -1 < 0 \)
\( f(\pi/6) = \frac{\pi-3}{6} > 0 \)
Therefore, since the the sign changes from negative to positive and the graph of f(x) is continuous on the given interval, by the Intermediate Value Theorem, the function has at least one x-intercept in the given interval, which is the same as saying that the equation has at least one root on the given interval.
September 26, 2024 | Quiz notes
P1
General feedback Remember to write neatly.
Please remember to use pencil.
Do not plug in on paper. If you would like to mentally plug in, use arrows pointing out of the expression.
Be sure to Justify!
a)
\( \lim\limits_{x\to -2}\frac{2-|x|}{2+x} = \lim\limits_{x\to -2}\frac{2-(-x)}{2+x} = \lim\limits_{x\to -2}\frac{2+x}{2+x} \)
-Because x is going to -2 and hence is less than zero, \( |x| = -x \) for both one-sided limits.
-The overall limit comes out to 1, after confirming both sides.
b)
\( \lim\limits_{x \to 0.5^{-}}\frac{2x-1}{x^2 |2x-1|} \)
Be sure to state that since \( x < 0.5 \), it follows that \( 2x-1 < 0 \), therefore \( |2x-1| = -(2x-1) \).
-The one sided limit, with work, comes out to -4
c)
\(\lim\limits_{h\to 0} \frac{(3+h)^{-1}-3^{-1}}{h} = \lim\limits_{h\to 0} (\frac{1}{3+h} - \frac{1}{3}) * \frac{1}{h} = \lim\limits_{h\to 0} \frac{3-(3+h)}{3 (3+h)} * \frac{1}{h} = \lim\limits_{h\to 0} \frac{-h}{3*h* (3+h)} = ... \)
-check that the parts of the fraction both go to zero.
-find the common denominator of the numerator
-simplify but keep writing the limit notation.
-the limit comes out to -1/9.
September 23, 2024 | Notes on today's quiz
If there are typos or mistakes, let me know so I can fix them here.
P1.
\( \lim\limits_{x\to -4} \frac{\sqrt{x^2+9}-5}{x+4} = ... = -4/5 \)
Check (with arrows) that the numerator and denominator both go to zero, then multiply both top and bottom by the conjugate of the numerator.
P2.
a) \( \lim\limits_{x\to 2^+} g(x) = \lim\limits_{x\to 2^+} \frac{x^2+x-6}{|x-2|} = \lim\limits_{x\to 2^+} \frac{(x-2)(x+3)}{x-2} = \lim\limits_{x\to 2^+} (x+3) = 5 \)
\( \lim\limits_{x\to 2^-} g(x) = \lim\limits_{x\to 2^-} \frac{x^2+x-6}{|x-2|} = \lim\limits_{x\to 2^-} \frac{(x-2)(x+3)}{-(x-2)} = \lim\limits_{x\to 2^-} (-x-3) = -5 \)
b) The overall limit \( \lim\limits_{x\to 2} g(x) \) does not exist but the one-sided limits are different.
c) Sketch a piece-wise function:
P3.
The statement is false. Consider the example we offered to illustrate the Squeeze Theorem:
\( \lim\limits_{x\to 0} (x^2 \sin(1/x)) = 0 \) despite the fact that \( \lim\limits_{x\to 0} (x^2) = 0 \) and \( \lim\limits_{x\to 0} \sin(1/x) \) does not exist.
September 23, 2024 | Notes on recent quiz
If there are typos or mistakes, let me know so I can fix them here.
Quiz 1 - P1
Draw the three pieces neatly. The
a values for which \( \lim_{a\to a} f(x) \) exists include all real numbers in the domain except for \( a= 0 \). The limit at this point does not exist because the one-sided limits are 1 and -1. At all other places, including \( a=1 \), the overall limit exists.
Quiz 1 - P2
Rewrite the equation as:
\( (e^{-x})^2-3 (e^{-x}) + 2 = 0 \)
Factor into: \( (e^{-x}-1) (e^{-x}-2) = 0 \)
\( e^{-x}=2 \text{ or } e^{-x} = 1 \)
Two solutions: \( x=-\ln 2, x=-\ln 1 = 0 \)
Quiz 2 - P1
No major issues with this problem. Be sure to label your graph. Put arrows on the coordinate axess.
Quiz 2 - P2
Factor completely: \( (\csc x - 2) (\csc x + 1) = 0 \)
Solve: \( \csc x = 2 \text{ or } \csc x = -1 \)
Find three solutions on the given domain: \( x= \pi/6, 5\pi/6, 3\pi/2 \)
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 13, 2024 | Notes on Today's quiz
If there are typos or mistakes, let me know so I can fix them here.
P1.
\( \frac{a}{24} = \sin\theta \rightarrow a = 24 \sin\theta \)
\( \frac{b}{24} = \cos\theta \rightarrow b = 24 \cos\theta \)
P2.
a) Set \( 9 - x^2 \geq 0 \) and solve either algebraically or graphically to find the domain: \( [-3, 3] \)
b) Set \( x^2 - 4 >0 \) and solve either algebraically or graphically to find the domain: \( x < -2 \cup x > 2 \)
c) The domain of the direct sum function is the intersection of the domains in parts a) and b).
The intersection, using interval notation, is: \( [-3, -2) \cup (2, 3] \).
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: 2MJA3Q
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.