notes / updates
October 14, 2025 | Quiz notes
P1.
\( f(x) = \frac{3x}{\sqrt{3x-1}} \rightarrow f'(x) = \frac{3 (3x-2)}{2 (3x-1)^{3/2}} \\
f'(x) = 0 \rightarrow x = \frac{2}{3}, f(2/3) = 2 \\
\left( \frac{2}{3},2\right) \)
P2.
\( f(x) = \frac{(1-x)^3}{x^2+2x} \rightarrow f'(x) = \frac{-(x-1)^2 (x^2+6x+2)}{(x^2+2x)^2} \\
f'(x) = 0 \rightarrow x=1, x=-3 \pm \sqrt{7} \)
October 9, 2025 | Quiz notes
P1.
\( f(x) = \cot x + 2x \rightarrow f'(x) = -\csc^2x +2 \)
The slope of the tangent line at \( x= \pi/6 \) is \( f'(\pi/6) = -2. \)
This means that the slope of the normal line is \( 1/2 \).
Use \( f(\pi/6) = \sqrt{3} + \frac{\pi}{3} \) and the point slope formula to find the equation of the normal line:
\( y = \frac{x}{2} + \sqrt{3} + \frac{\pi}{4} \)
P2.
\( y = \tan x \rightarrow \frac{dy}{dx} = \sec^2 x = 2 \). Give a reason why the slope has to equal 2: because we are looking for a tangent line that is parallel to the given line.
Solve the equation in the given domain, to find: \( x = \pm \frac{\pi}{4} \).
Be sure to compute and include the y-coordinates in your answer: \( (\frac{-\pi}{4}, -1), (\frac{\pi}{4},1) \).
October 5, 2025 | Feedback on recent quiz
The definition of slope as a number is either:
\( m = \lim\limits_{h \to 0}\frac{f(-3+h)-f(-3)}{h} \)
OR
\( m = \lim\limits_{x \to -3}\frac{f(x)-f(-3)}{x-(-3)} \)
Choose one and see where it takes you. Below I show work with the second definition:
\( m = \lim\limits_{x \to -3}\frac{f(x)-f(-3)}{x-(-3)} = \lim\limits_{x \to -3}\left(\frac{12}{3-x}-\frac{12}{3-(-3)}\right) \frac{1}{x+3} = \)
\( = 12 \lim\limits_{x \to -3}\left(\frac{1}{3-x}-\frac{1}{6}\right) \frac{1}{x+3} = 12 \lim\limits_{x \to -3}\left(\frac{6-(3-x)}{6(3-x)}\frac{1}{x+3}\right) = \)
\( = 12 \lim\limits_{x \to -3}\left(\frac{3+x}{6(3-x)}\frac{1}{x+3}\right) = 12\lim\limits_{x \to -3}\left(\frac{1}{6(3-x)}\right) = \frac{12}{36} = \frac{1}{3} \)
Since \( f(-3) = \frac{12}{3-(-3)} = 2 \), we use the point-slope formula to find the equation of the tangent line:
\( y - 2 = \frac{1}{3} (x-(-3)) \), which simplifies to \( y = \frac{1}{3} x + 3\)
September 28, 2025 | Feedback on recent quiz
P1. a)
\( \lim\limits_{x\to \pi^-} 2^{\cot h} = \lim\limits_{x\to \pi^-} 2^{\frac{\cos h}{\sin h}} \)
Use arrows or explain how cosine goes to -1 and sine goes to 0 from positive side, so the cotangent expression goes to negative infinity.
The overall limit is therefore 0, since \( 2^{-\infty} = \frac{1}{2^{\infty}} \to 0 \)
P1. b)
\( \lim\limits_{\theta \to 0} \sec\left( \frac{1-\cos(2\theta)}{\theta} \right) = \sec \left( \lim\limits_{\theta \to 0} \frac{1-\cos 2\theta}{\theta} \right) = \\
\sec \left(2 \lim\limits_{\theta \to 0} \frac{1-\cos 2\theta}{2 \theta} \right) \)
The inner expression goes to 0, so the overall limit is \( \sec (2*0) = \sec(0) = 1 \)
P1. c)
\( \lim\limits_{t \to 0^+} \left( \ln(\tan t) -\ln(\sin 2t) \right) = \lim\limits_{t \to 0^+} \ln \left[ \frac{\tan t}{\sin 2t} \right] = \\
=\ln \lim\limits_{t \to 0^+} \left[ \frac{\sin t}{2 \cos^2 t \sin t} \right] = \ln(1/2) = -\ln 2 \)
P2.
Let the left side of the equation be the function \( f(x) = x + \sin x - 1 \). Note that since all three terms are continuous for all real numbers, the sum is also continuous everywhere.
Therefore, \( f(x) \) is continuous on the interval \( [0, \frac{\pi}{6}] \), which means we can apply the Intermediate Value Theorem.
Note that \( f(0) = -1 \) and \( f(\pi/6) =\frac{\pi}{6} + \frac{1}{2} - 1 \approxeq 0.52 - 0.5 > 0\).
Since the function changes sign and is continuous on the given interval, by the Intermediate Value Theorem, it must have at least one zero on the given interval.
This is the same as saying: the equation \( f(x) = 0 \) has at least one solution on the interval \([0, \frac{\pi}{6}] \).
September 23, 2025 | Working with Lists - TI NSPIRE
For those of you who own an NSPIRE, feel free to use the following tutorial with text and graphics.
Stop after the section where it says "Selecting a Column (or Row)".
Link:
TI NSPIRE TUTORIAL
September 22, 2025 | Notes on today's quiz
If there are typos or mistakes, let me know so I can fix them here.
P1.
\( \lim\limits_{x\to-4^+} g(x) =\lim\limits_{x\to-4^+} \frac{(x+1)(x+4)}{|x+4|} = \lim\limits_{x\to-4^+} \frac{(x+1)(x+4)}{x+4} = \lim\limits_{x\to-4^+} (x+1) = -3 \)
\( \lim\limits_{x\to-4^-} g(x) =\lim\limits_{x\to-4^-} \frac{(x+1)(x+4)}{|x+4|} = \lim\limits_{x\to-4^-} \frac{(x+1)(x+4)}{-(x+4)} = \lim\limits_{x\to-4^-} -(x+1) = 3 \)
The overall limit does not exist since the one-sided limits are different.
The graph of the function consists of two linear pieces:
\( g(x) = x+1 \) for \( x > -4 \).
\( g(x) = -x-1 \) for \( x < -4 \).
A desmos generated graph is below:
P2.
Check that the numerator and denominator both go to zero, then use the conjugate:
\( \lim\limits_{x\to -6} \frac{\sqrt{x^2+64}-10}{x+6} = \lim\limits_{x\to -6} \frac{\sqrt{x^2+64}-10}{x+6} \frac{\sqrt{x^2+64}+10}{\sqrt{x^2+64}+10} \\ = \lim\limits_{x\to -6} \frac{x^2-36}{x+6} \frac{1}{\sqrt{x^2+64}+10} =\lim\limits_{x\to -6} \frac{(x-6)(x+6)}{x+6} \frac{1}{\sqrt{x^2+64}+10} = \lim\limits_{x\to -6} \frac{x^2-36}{x+6} \frac{1}{\sqrt{x^2+64}+10} =\lim\limits_{x\to -6} \frac{x-6}{\sqrt{x^2+64}+10} = \frac{-12}{20} = \frac{-3}{5} \)
P3.
The statement is false. As an example, consider: \( f(x) =\frac{|x|}{x}, g(x) = \frac{-|x|}{x} \)
The individual limits of these two functions as \(x\) approaches zero do not exist. The limit of the product is -1, however.
September 18, 2025 | Notes on the recent quiz
If there are typos or mistakes, let me know so I can fix them here.
P1.
The sketching problem went well on average. Some of the papers had the curves that were not functions (they failed the vertical line test)
Also, remember to draw neatly and
approach the asymptote without crossing it. It's okay to cross the horizontal asymptote as long as you make sure the graph
returns towards it.
P2.
The sketch of the piece-wise function is shown below with desmos. The limit exists for all real values except \( a = 0 \).
The one sided limits as \( x \) approaches zero are 1 and -1, which means \( \lim\limits_{x\to 0} f(x) \) does not exist.
September 12, 2025 | Notes on Today's quiz
If there are typos or mistakes, let me know so I can fix them here.
P1.
a) Set \( 9 - x^2 \geq 0 \) and solve either algebraically or graphically to find the domain: \( [-3, 3] \)
Algebraically, study the sign after factoring \( (3-x)(3+x) \geq 0 \).
Geometrically, draw the graph of \( y = 9 - x^2 \) to confirm the domain.
b) Set \( x^2 - 4 >0 \) and solve either algebraically or graphically to find the domain. The work is similar to part a).
The domain is: \( x < -2 \cup x > 2 \). In interval notation: \( (-\infty, -2,) \cup (2, \infty) \)
c) Set \((1-x)(5-x) > 0 \) and solve either algebrically or graphically to find the domain. The work is similar to part a).
The domain is: \( (1,5) \).
c) The domain of the direct sum of the three functions is the intersection of the domains in parts a), b), and c).
The intersection, using interval notation, is: \( (2, 3] \).
September 8, 2025
AP registration instructions at this [
link ].
If you think you'll be changing your schedule or dropping the class altogether, please hold off on joining the AP classroom.
For your course section, use join code: QELZR2
If you have any questions, contact Mr. Dan Monahan (AP Coordinator).
September 1, 2025 | Introductory note.
Welcome to AP Calculus BC. In this area of the course page we will post important course-related information such as links to resources, announcements, reminders, commentary or feedback on assessments etc. All entries will be dated and the most recent one will be at the top of the page. As far as daily course information goes, the Assignments and Bulletin tabs are the most important.