notes / updates
November 19, 2024 | Thoughts on recent quiz
If there are typos or mistakes, let me know so I can fix them here.
Vocab Quiz:
a) The Second Derivative Test is used to justify local extrema. Be sure to state that f(x) has a stationary number at x=c, so the first derivative is zero there. List all three cases about the second derivative:
If the second derivative at x=c is positive, then f(c) is a local minimum.
If the second derivative at x=c is negative, then f(c) is a local maximum.
If the second derivative is also zero, then the test is inconclusive.
b) EVT states that a function f(x) that is continuous on a closed interval [a, b] is guaranteed to have an absolute maximum and an absolute minimum on that interval.
c) x=c is a critical number of a function f(x) if f'(c) is either zero or undefined.
d) A function f(x) has an inflection point at x=c if three conditions are met: f is continuous, f'(x) does NOT change sign at x=c, and f''(x) changes sign at x=c.
e) The First Derivative Test is used to justify local extrema. Be sure to state the function must be continuous at a critical number x=c.
Case 1: If f' changes sign from + to - at x=c, then f(c) is a local maximum.
Case 2: If f' changes sign from - to + at x=c, then f(c) is a local minimum.
Case 3: If f' does not change sign at x=c, then f(c) is not a local extremum.
Curve-Sketching Quiz.
On average, this task went well. A few points:
For the Slant Asymptote be sure to divide the polynomials. Don't just list the answer \( y = x \).
For symmetry, be sure to verify that it is odd: \( f(-x) = -f(x) \) for all \( x \).
The fully factored derivatives were: \( f'(x) = \frac{x^2 (x^2+3)}{(x^2+1)^2}; f''(x) = \frac{-2x (x-\sqrt{3})(x+\sqrt{3})}{(x^2+1)^3} \)
Be sure to list or label the tree inflection points, which means we include their y-coordinates as well.
Be sure to state that there are no extrema of any kind.
November 18, 2024 | Resource: WolframAlpha
WolframAlpha is a reliable resource that you can use to confirm various mathematical calculations. To get started, visit wolframalpha.com and type a query using natural language.
For example:
derivative of cos(3x) at x equals pi/6 . The following url gives the results:
https://www.wolframalpha.com/input?i=derivative+of+cos%283x%29+at+x+equals+pi%2F6
November 15, 2024 | Thoughts on recent quiz
If there are typos or mistakes, let me know so I can fix them here.
P1.
\( f(x) = x - \frac{1}{x}-\frac{1}{x^2} = \frac{x^3-x^2-1}{x^2} \)
Domain: all reals except \( x = 0 \).
Symmetry: none.
VA: \( x=0 \)
HA: none.
Slant: \( y=x \)
y-intercept: none
x-intercept: at least one due to IVT but it's hard to find since it is difficult to solve: \( x^3-x^2-1 = 0 \)
\( f'(x) = \frac{(x+1)(x^2-x+2)}{x^3} \rightarrow x=-1, x=0\) are the critical numbers.
The sign table for \( f'(x) \) only needs 2 factor rows because the quadratic expression is always positive \( b^2-4ac < 0, a>0 \)
ID behavior: increasing on \( (-\infty, -1) \cup (0, \infty) \); decreasing on \( (-1,0) \)
\( f''(x) = \frac{-2(x+3)}{x^4} \rightarrow x=-3, x=0 \) are the critical numbers.
The sign table for \( f''(x) \) also needs two factor rows.
Concavity: concave up on \( (-\infty, -3)\); concave down on \( (-3, 0) \cup (0, \infty) \)
Inflection point at: \( (-3, -25/9) \); Local maximum at \( (-1, -1) \) ; No other extrema.
Conclude with sketch. A version from desmos with key details is below:
P2.
\( f(x) = \frac{x^2}{\sqrt{x+1}} \)
Domain: \( x > -1 \).
Symmetry: none.
VA: \( x= -1 \)
HA: none.
Slant: none
y-intercept: \( (0,0) \)
x-intercept:\( (0,0)\)
\( f'(x) = \frac{x(3x+4)}{2(x+1)^{3/2}} \rightarrow x=0, x=-1\) are the only critical numbers in the domain.
The sign table for \( f'(x) \) needs 3 factor rows or only 2 if you make a note about \( (x+1)^{3/2} \) being positive for \( x >-1 \)
ID behavior: decreasing on \( (-1, 0) \); increasing on \( (0,\infty) \)
\( f''(x) = \frac{3x^2+8x+8}{4(1+x)^{5/2}} \rightarrow \) no critical numbers in the domain.
You do not need a sign table for \( f''(x) \) as long as you state that both numerator and denominator are positive wherever they are defined.
Concavity: concave up on \( (-1, \infty )\)
No inflection points. Local and global minimum at \( (0, 0) \) ; No other extrema.
Conclude with sketch. A version from desmos with key details is below.
P3.
\( f(x) = e^{(-x^2)/2} \)
Domain: all reals.
Symmetry: even.
VA: none
HA: \( y = 0 \) with brief explanation (use limits)
Slant: none
y-intercept: \( (0, 1) \)
x-intercept: none
\( f'(x) = -x e^{(-x^2)/2} \rightarrow x=0\) is the only critical number.
The sign table for \( f'(x) \) only needs 1 factor row so we don't even need a table. A number line will suffice here.
ID behavior: increasing on \( (-\infty, 0) \); decreasing on \( (0,\infty) \)
\( f''(x) = (x-1)(x+1) e^{(-x^2)/2} \rightarrow x=-1, x=1 \) are the critical numbers.
The sign table for \( f''(x) \) needs two factor rows.
Concavity: concave up on \( (-\infty, -1) \cup (1, \infty) \); concave down on \( (-1, 1) \)
Inflection points at: \( \left(-1, \frac{1}{\sqrt{e}}\right), \left(1, \frac{1}{\sqrt{e}}\right) \); Local and global maximum at \( (0, 1) \) ; No other extrema.
Conclude with sketch. A version from desmos with key details is below.
P4.
\( f(x) = \frac{\cos x}{2+\sin x}\)
Domain: \( [0, 2\pi] \), as given.
Symmetry: none.
VA: none
HA: none or NA due to the restricted domain.
Slant: none or NA due to restricted domain.
x-intercepts: \( (\pi/2, 0), (3\pi/2, 0) \)
y-intercept: \( (0, 1/2) \)
\( f'(x) = \frac{-2\sin x - 1}{(2+\sin x)^2} \rightarrow x=7\pi/6, x=11\pi/6\) are the critical numbers.
The sign table for \( f'(x) \) only needs 1 factor row (for the numerator!) so as long as we explain, a number line with explanation would be sufficient here.
ID behavior: increasing on \( (7\pi/6, 11\pi/6) \); decreasing on \( (0,7\pi/6) \cup (11\pi/6, 2\pi) \)
\( f''(x) = \frac{2 \cos x (\sin x - 1)}{(2+\sin x)^3} \rightarrow x=\pi/2, x=3\pi/2 \) are the critical numbers.
The sign table for \( f''(x) \) needs two factor rows, one for each expression at the numerator level.
Concavity: concave up on \( (\pi/2, 3\pi/2) \); concave down on \( (0, \pi/2) \cup (3\pi/2, 2\pi) \)
Inflection points at: \( (\pi/2, 0), (3\pi/2, 0) \); Local and global maximum at \( (11\pi/6, \frac{\sqrt{3}}{3}) \) ; Local and global minimum at \( (7\pi/6, -\frac{\sqrt{3}}{3}) \)
Conclude with sketch. A version from desmos with key details is below.
November 14, 2024 | Thoughts on recent quiz
If there are typos or mistakes, let me know so I can fix them here.
P1.
The function \( f(x) = x^3 e^{-2x}\) is a product of two continuous functions on the entire domain of real numbers, therefore it is continuous on the interval [1, 4].
We use the Closed Interval Method here. (Be sure to state the name of the method and the reason why the function is continuous.)
Find the derivative and consider two cases: when/if it equals 0 and when/if it is undefined.
\( f'(x) = \frac{x^2 (3-2x)}{e^{2x}} \rightarrow x=0, x=3/2\)
Of the two critical numbers, only 3/2 is in the given domain. Evaluate \( f(3/2) = \frac{27}{8e^3}, f(4) = \frac{64}{e^8}, f(1) = \frac{1}{e^2} \)
to find that the absolute maximum is \( \frac{27}{8e^3} \) and the absolute minimum is \( \frac{64}{e^8} \).
November 11, 2024 | Thoughts on recent quiz
If there are typos or mistakes, let me know so I can fix them here.
P1.
a) Use L'Hospital's Rule twice. Be sure to use arrows to indicate that the conditions have been confirmed.
\( \lim\limits_{p \to 0} \frac{0.5p^2+\cos p}{p^4} = \frac{1}{24}\)
b) Confirm that this is an indeterminate form of the type \( 1^{\infty} \), then proceed with:
\( \ln y = bx \ln\left(1+ \frac{a}{x} \right) \)
Find the limit of the original expression to be \( e^{ab} \) since \( \ln y \rightarrow ab \) as \( x \to \infty\).
c) Check the conditions before rewriting as: \( \lim\limits_{x\to \infty} \frac{\tan(1/x)}{1/x}\).
The limits ends up being 1 here.
November 5, 2024 | Thoughts on recent quiz
If there are typos or mistakes, let me know so I can fix them here.
P1.
Suppose the tangent line touches the curve at \( x = a \)
Find slope after evaluating the derivative: \( f'(x)=3e^{3x} \rightarrow f'(a) = 3e^{3a} \)
The slope is also equal to the slope of the segment connecting the origin to the point on the curve: \( \frac{e^{3a}-0}{a-0} = \frac{e^3a}{a} \)
Set the two versions of slope equal to each other, and then solve to find: \( a=\frac{1}{3} \)
The point on the curve is: \( (\frac{1}{3}, e) \).
P2.
\( L(x) = f(\pi/3) + f'(\pi/3) (x-\pi/3) \rightarrow L(x) = \sqrt{3} + 4(x-\pi/3) \)
Plugging in \( x = 61\pi/180 \), we get:
\( \tan(61^{\circ}) \approxeq L(61\pi/180) = \sqrt{3} + \pi/45 \)
P3.
Draw a right triangle with hypotenuse \( c \) and legs \( a, b \).
We know that: \( \frac{dc}{dt} = m, \frac{da}{dt} = -n \)
Let \( \alpha \) be the angle formed by \( c \) and \( b \).
For part a) rewrite the relationship: \( \sin\alpha = \frac{a}{c} \) as \( \alpha = \arcsin(\frac{a(t)}{c(t)}) \)
For part b) differentiate with respect to time: \( \frac{d\alpha}{dt} = \frac{1}{\sqrt{1-\frac{a^2}{c^2}}} \frac{c*da/dt - a*dc/dt}{c^2} \)
Plug in the known rates as well as \( c = \sqrt{2} \) to find \( \frac{d\alpha}{dt} = -n-\frac{m}{\sqrt{2}} \) radians per second.
At the instant when both legs are 1 centimeter, the angle \( \alpha \) formed by the hypotenuse and side \( b \) decreases at a rate of \( n+ \frac{m}{\sqrt{2}} \) radians per second.
November 4, 2024 | Thoughts on recent quiz
If there are typos or mistakes, let me know so I can fix them here.
P1.
\( f(x) = \sin^{-1}\frac{x}{2} \) The y-value is 0 when \( x= 0\), so the tangent line touches the curve at the origin.
\( dy/dx = \frac{1}{\sqrt{1-\frac{x^2}{4}}}\frac{1}{2} \) by the chain rule.
Evaluate at the origin to get \( f'(0)=1/2 \)
The tangent line equation is \( y= x/2 \)
P2.
Use logarithmic differentiation to find \( m = 3 \) and the tangent line equation is \( y=3x. \)
The first steps are below:
\( y = \frac{x (\sqrt[3]{x}+1)^2}{\sqrt{x+1}} \)
\( \ln y = \ln x + 2 \ln(\sqrt[3]{x}+1) -\frac{1}{2} \ln(x+1) \)
\( \frac{1}{y}\frac{dy}{dx} = \frac{1}{x} + \frac{2}{\sqrt[3]{x}+1} \frac{1}{3}*x^{-2/3} -\frac{1}{2(x+1)} \)
October 25, 2024 | Thoughts on recent quiz
If there are typos or mistakes, let me know so I can fix them here.
P1.
This problem went well. Be sure to answer the question:
At the moment when \( x=1, y=2 \) the quantity \( z \) is changing at a rate of \( -12 \) units per second.
The explanation: Since the rate is negative, \( z \) is decreasing at that moment.
P2.
No major issues here either. When the area is 9 square miles, the radius of the spill is increasing at a rate of \( \frac{1}{\sqrt{\pi}} \) miles per hour.
Be sure to simplify. If you write the rate as \( \frac{\sqrt{\pi}}{\pi} \), that is fine, too.
October 23, 2024 | Thoughts on recent quiz
If there are typos or mistakes, let me know so I can fix them here.
P1.
Problem 1 generally went well. This was nearly identical to the question we did in class. For the cone, be sure to draw neatly and set of the ratio of similar triangles:
\( \frac{r}{3} = \frac{h}{10} \rightarrow r=\frac{3h}{10} \)
Substitute this into the volume formula so that the volume is in terms of \( h(t) \) only.
If you take the derivative implicitly with respect to time without substituting, be sure to use:
\( \frac{dr}{dt} = \frac{3}{10} \frac{dh}{dt} \)
The unknown rate was: \( \frac{dh}{dt} = \frac{8}{9\pi} cm/s \)
Be sure to include a sentence for the conclusion.
P2.
Be sure to start with a correct drawing. If \( y \) is the length of the shadow and \( x \) is the distance from the spotlight, the ratio of similar triangles gives us:
\( \frac{ 12}{y} = \frac{x}{2} \)
Rewrite as: \( 24 = x(t) y(t) \) and then implicitly differentiate with respect to time \( t \)
\( \frac{dy}{dt} = \frac{-3}{5} m/s \). Be sure to include a sentence for the conclusion.
October 21, 2024 | Thoughts on recent quiz
If there are typos or mistakes, let me know so I can fix them here.
P1.
The equation of the curve is: \( x y + 2x-y=0 \)
Read the slope to the curve as: \( m=1/2 \) (the negative reciprocal of the slope of the given line)
Differentiate completely: \( y + x \frac{dy}{dx} + 2 - \frac{dy}{dx} = 0 \) and plug in \( \frac{dy}{dx}=\frac{1}{2} \) to get: \( x=-2y-3 \)
Plug this back into the equation of the curve, and solve for: \(y = -1, y=-3 \)
Find the x-values, so that the points are: \( (-1, -1) \text{ and } (3, -3) \)
The normal line equations are: \( y+1 = -2 (x+1) \text{ and } y +3 = (-2)(x-3) \)
The graph below shows the curve, the given line, and the two tangent lines whose equations we found:
P2.
The equation of the curve is: \( x^2+ xy + y^2 = 7 \)
Setting y =0, we get: \( x= \pm \sqrt{7} \) (show work)
Differentiate implicitly: \( 2x + x \frac{dy}{dx} + y + 2 y \frac{dy}{dx} = 0 \), then solve:
\( \frac{dy}{dx}=\frac{-2x-y}{x+2y} \)
Plug in each of \( (\pm\sqrt{7}, 0) \) to get \( \frac{dy}{dx} = -2 \) at both points.
In case you are curious, the graph below shows the curve, the given line, and the two parallel tangent lines at the points we found:
October 14, 2024 | 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 8, 2024 | Quiz notes
P1
Use the quotient rule. Do NOT use techniques we have not discussed in class.
\( f(x) = \frac{2}{1-3x} \rightarrow f'(x) = \frac{(1-3x)*0 - (-3)*2}{(1-3x)^2} = \frac{6}{(1-3x)^2} \)
When \(x = 0, y = 2, f'(0) = 6 \rightarrow y = 6x + 2\)
When \(x = -1, y = \frac{1}{2}, f'(-1) = \frac{3}{8} \rightarrow y = \frac{3}{8}x + \frac{7}{8}\)
P2.
The slope of the given line is \(-3 \), since the equation can be written as \( y = -3x \).
Find the derivative using the power rule or the quotient rule:
\( y = 12 x^{-1} \rightarrow \frac{dy}{dx} = -12 x^{-2} = \frac{-12}{x^2} \)
Set the derivative equal to \( -3 \), then solve to find \( x= \pm 2 \).
The points are \( (2, 6) \) and \( (-2,-6) \).
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.