notes / updates
November 21, 2024 | Thoughts on recent quiz
If there are typos or mistakes, let me know so I can fix them here.
P1.
\( f(x) = \arctan(-4x) -\ln(2x)\) at \(x=\frac{1}{4} \)
\( f(1/4) = \ln 2 - \frac{\pi}{4} \)
\( f'(x) = \frac{-4}{1+16x^2} -\frac{1}{x}\)
The slope of the tangent line is therefore: \( f'(1/4) = -6\)
The tangent line equation is: \( y = -6x + \frac{3}{2} +\ln 2 -\frac{\pi}{4}\)
The normal line equation is: \( y = \frac{x}{6} - \frac{1}{24} +\ln 2 -\frac{\pi}{4}\)
P1.
The point (1,2) is on the graph of \( g(x) \) because the tangent line touches there.
Since the slope of the tangent line is 3, that means \( g'(1)= 3\)
The point (2,1) is on the graph of \( g^{-1}(x) \).
So we have: \( (g^{-1})'(2) = \frac{1}{g'(g^{-1}(2))} = \frac{1}{g'(1)} = \frac{1}{3} \)
The tangent line on \( g^{-1}\) is then: \( y- 1 = \frac{1}{3} (x-2) \) or \( y =\frac{x+1}{3}\).
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.
\( y = \frac{x^3 \sqrt[3]{x^2+7}}{(3-x)^6} \)
When \( x= 1 \), find \( y=\frac{1}{32} \).
In order to find the slope of the tangent line, we first take the natural log of both sides:
\( \ln y = 3 \ln x -\frac{1}{3} \ln(x^2+7) - 6\ln(3-x) \)
Then we differentiate implictly (i.e. logarithmic differentiation):
\( \frac{1}{y} \frac{dy}{dx} = \left( \frac{3}{x} +\frac{1}{3} \frac{2x}{x^2+7} - 6 \frac{-1}{3-x} \right) \)
Plug in \( x=1, y=\frac{1}{32} \) to find \( m = \frac{73}{384}\)
The equation of the tangent line is: \( y -\frac{1}{32} = \frac{73}{384}(x-1) \) or \( y = \frac{73x-63}{384}\)
Note: the quotient rule for this problem is totally fine but it will turn out to be messy and hence we are more likely to make mistakes if we use that approach.
P2.
\( f(x) = \frac{1}{3} (x^5+2x^3)\)
By definition, \( (f^{-1}(a))' = \frac{1}{f'(f^{-1}(a))}\).
When \( a=-1 \), the statement above becomes: \( (f^{-1}(-1))' = \frac{1}{f'(f^{-1}(-1))}\)
Use precalculus to find \( f^{-1}(-1) \) by setting \(y=1 \) in the original function: \( -1 = \frac{1}{3} (x^5+2x^3) \)
Rewrite as \( -3 = x^5+ 2x^3 \) and by trial and error confirm that \( x = -1 \), which means \( f^{-1}(-1)= -1\) (A coincidence? Maybe.)
\( (f^{-1}(-1))' = \frac{1}{f'(-1)} =\frac{1}{\frac{1}{3}(5(-1)^4+6*(-1)^2)} = \frac{3}{5+6} = \frac{3}{11}\)
November 14, 2024 | Thoughts on recent quiz
If there are typos or mistakes, let me know so I can fix them here.
P1.
No need for a drawing here.
Use the product rule when fully differentiating: \( V = I R \rightarrow \frac{dV}{dt} = I \frac{dR}{dt} + R \frac{dI}{dt}\)
Plug in the given information to find the unknown rate with correct units: \( \frac{dI}{dt} = \frac{-3}{4}\) amperes per second.
Conclusion: When the voltage is 12 volts and the resistance is 4 ohms, the current is changing at a rate of \( \frac{-3}{4}\) amperes per second.
P2.
Make a drawing. Sadly, some of us struggled with the details here.
If \( z(t) \) is the distance from the observer to the car and \( x(t) \) is the distance from the car to the intersection once it has passed it, the equation becomes:
\( x^2 + 50^2 = z^2 \)
Note that 5 seconds later the values are \( x = 225 m, z = 25 \sqrt{85} m\)
The unknown rate \( \frac{dz}{dt} \) comes out to \( \frac{405}{\sqrt{85}} m/s \)
Be sure to include a full and grammatically sentence for your conclusion.
P3.
-The question is about a conical tank. Do not draw a cylinder. Do not draw a cone with the vertex at the top.
-Draw a big cone (tank) and an inner cone (water)
-Find \( r = \frac{5h}{8}\) by using similar triangle.
-The equation becomes \( V= \frac{25\pi}{192} h^3 \)
-When the tank is half full, the height is \( h = 4\sqrt[3]{4}\)
-The unknown rate is \( \frac{dh}{dt} = \frac{6}{125\pi \sqrt[3]{2}}\) ft/hr.
-Be sure to include a sentence.
P4.
-Draw a triangle with hypotenuse equal to 5cm. If \( x \) is the side opposite the acute angle, the equation becomes: \( \frac{x}{5} = \sin\theta\)
-Find \( \frac{dx}{dt} = 12\) cm/min.
-When \( x = 3 \), the value of cosine is 4/5. (A right triangle with sides 3, 4, 5.)
-Include a sentence.
November 11, 2024 | Thoughts on recent quiz
If there are typos or mistakes, let me know so I can fix them here.
P1.
Make a drawing of the ground level, a fixed point where the telescope is and a plane at an altitude of 5km, somewhere to the right.
If \(\theta \) is the angle of elevation, \( x \) is the adjacent, and 5 km is the side opposite the angle \( \theta \), we have:
\( \tan\theta = \frac{5}{x} \) which can also be written as: \( \cot\theta = \frac{x}{5} \)
After differentiating both sides with respect to time, plug in \( x= \frac{5}{\sqrt{3}}, \frac{d\theta}{dt} = \frac{-\pi}{6} \)
The final answer is: \( \frac{dx}{dt} = \frac{10\pi}{9} km/min \)
Be sure to include a conclusion with the final answer and correct units
P2.
Draw a shoreline, a lighthouse 2000 meters away, and a point on the shoreline that is x meters away from the point on the shore nearest to the lighthouse.
Form a right triangle by making the hypotenuse the segment that connects the lighthouse with the point down the shoreline. The angle \(\theta \) is the angle the beam of light forms with the leg that connects the lighthouse to the nearest point on the shore.
The "big" equation is: \( \tan\theta = \frac{x}{2000} \)
After differentiating with respect to time, plug in: \( \sec^2\theta = \frac{17}{16} , \frac{d\theta}{dt} = 4\pi \) to find: \( \frac{dx}{dt} = 8500\pi m/min\).
Be sure to write a conclusion with the final answer and correct units.
November 7, 2024 | Quiz Feedback
Below find some notes & feedback on the recent quiz.
P1.
\( 6 + x^3y = xy^2 \)
a) Differentiate with respect to \(x \). Be sure to use Leibniz notation, as follows. (Do not use \( y'\)!)
\( 0 + 3x^2 y +x^3 \frac{dy}{dx} = y^2 + 2xy \frac{dy}{dx}\)
Find \( \frac{dy}{dx} = \frac{3x^2y-y^2}{2xy-x^3} \) or \( \frac{dy}{dx} = \frac{y^2-3x^2y}{x^3-2xy} \)
b) When \( x = 1 \), the equation of the curve becomes \( 6+y = y^2 \)
Solve the quadratic equation to find two solutions: \(y=3, y=-2\), so the points are \( (1,3) \) and \( (1,-2)\).
At \( (1,3)\), the slope is zero, so the tangent line has equation \( y =3 \).
At \( (1,-2)\), the slope is 2, so the tangent line has equation \( y=2x-4\).
c) The slope must be undefined when the tangent line is vertical, so be sure to set the derivative's denominator equal to zero:
\( 2xy - x^3 = 0 \)
\( x (2y-x^2) = 0 \)
Case 1: \( x = 0\). Plugging this into the original equation gives us 6 = 0, which is an impossibility.
Case 2: \( y = \frac{x^2}{2} \). Plugging this into the original equation gives us \( x =\sqrt[5]{-24}\), which is the equation of the vertical tangent line.
November 4, 2024 | Quiz Feedback
Below find some notes & feedback on the recent quiz.
P1.
\( f(x) = x +\cot x \)
At \( x=\frac{\pi}{6}, f(\pi/6) = \frac{\pi}{6} + \sqrt{3} \)
The slope of the tangent line can be determined by using the derivative:
\( f'(x) = 1-\csc^2x \rightarrow f'(\pi/6) = 1-4= -3 \)
The slope of the normal line is therefore the negative reciprocal, which is \( \frac{1}{3} \)
The equation of the normal line is: \( y =\frac{1}{3} x +\frac{\pi}{9} +\sqrt{3} \)
P2.
-Make a drawing and set up the equation: \( \sin\beta = \frac{\text{opposite}}{\text{hypotenuse}} = \frac{x}{20} \)
-Optionally, you're welcome to rewrite as: \( x = 20 \sin\beta \)
-Find the derivative: \( \frac{dx}{d\beta} = 20 \cos\beta \)
-Plug in \( \beta = \frac{\pi}{6} \) to find: \( \frac{dx}{d\beta} = 10\sqrt{3} \) feet per radian.
November 1, 2024 | Quiz Feedback
Below find some notes & feedback on the recent quiz.
P1.
\( f(x) = (2x-1)\sqrt{4x+1} \)
(2): Find the derivative in its simplified form: \( f'(x) =\frac{12x}{\sqrt{4x+1}} \)
(1): Set the derivative equal to zero in order to find points where the tangent line is horizontal.
(1): Find \( x=0 \)
(1): Since it asks for points, we need to find the y-coordinate as well: \( f(0)=-1 \), so the point is \( (0,-1) \)
(1): For part b) find the y-value when \( x=2 \), so \( f(2) = 9 \)
(1): Find the slope at this point: \( f'(2) = 8 \)
(1): Find and simplify(!) the equation of the tangent line: \( y=8x-7\)
October 31, 2024 | Quiz Feedback
Below find some notes & feedback on the recent quiz.
P1.
\( y =3x - \tan x \)
(1) Read the slope of the given line as 1 after rewriting the equation as \(y = x+2 \)
(1) Find the derivative: \( \frac{dy}{dx} = 3-\sec^2 x \)
(1) State that for the lines to be parallel the derivative must equal the slope of the given line, so \( \frac{dy}{dx} = 3-\sec^2x = 1 \)
(1) Transform the condition into: \( \sec^2 x = 2 \)
(1) Find four solutions, one in each quadrant: \( \cos x = \pm \frac{\sqrt{2}}{2} \rightarrow x= \pi/4, 3\pi/4, 5\pi/4, 7\pi/4 \)
(1) State the general solutions: \( x = \frac{\pi}{4} + k\frac{\pi}{2} \)
October 24, 2024 | Quiz Feedback
Below find some notes & feedback on the recent quiz.
P1.
\( f(x) = 2x \sin x \rightarrow f'(x)=2\sin x + 2x \cos x \)
\( f(\pi/2) = \pi, f'(\pi/2) = 2 \)
The tangent line has equation: \( y-\pi = 2 (x-\pi/2) \), which simplifies to: \( y=2x \).
P2.
\( y = 3x + 6\cos x \rightarrow \frac{dy}{dx} = 3 - 6\sin x \)
Plug in \( x=\pi/3 \) into the derivative to find slope: \( m = 3- 3\sqrt{3} \)
Plug in \( x=\pi/3 \) into the original equation to find the y-value: \( y_1 = \pi + 3 \)
The equation of the tangent line is: \( y-(\pi + 3) = (3-3\sqrt{3}) (x-\pi/3) \).
This simplifies to: \( y = (3-3\sqrt{3}) x + 3 + \sqrt{3} \pi \).
October 18, 2024 | Quiz Feedback
Below find some notes & feedback on the recent quiz.
Friendly reminders:
Read and follow instructions.
Write neatly.
Do not use techniques we have not covered in class.
P1.
\( f(x) = \frac{2}{1-3x} \)
For the
derivative as a number use one of the following:
\( f'(-1)=\lim\limits_{h\to 0} \frac{f(-1+h)-f(-1)}{h} \) or \( f'(-1) = \lim\limits_{x\to -1} \frac{f(x)-f(-1)}{x-(-1)} \)
The question did NOT (!) ask to use the
definition of the derivative as a function: \( \lim\limits_{h\to 0}\frac{f(x+h)-f(x)}{h} \).
The slope comes out to \( \frac{3}{8} \) and the equation of the tangent line is \( y= \frac{3}{8}x + \frac{7}{8} \).
P2.
\( f(x) = \frac{4x}{x^2+1} \)
Use the quotient rule:
\( f'(x) = \frac{(x^2+1)*4 - 4x*(2x)}{(1+x^2)^2} = \frac{4(1-x^2)}{(x^2+1)^2} \)
At (0,0), the slope is 4, so the tangent line is \( y =4x\).
At (1,2), the slope is 0, so the tangent line is \( y =2\).
October 17, 2024 | Quiz Feedback
Below find some notes & feedback on the recent quiz.
First Quiz
\( f(x) = \frac{1}{\sqrt{x}} \)
\( f'(x)=\lim\limits_{h\to 0} \frac{f(x+h)-f(x)}{h} \)
\( = \lim\limits_{h\to 0} \frac{\frac{1}{\sqrt{x+h}}-\frac{1}{\sqrt{x}}}{h} \)
\( = \lim\limits_{h\to 0} \frac{1}{h} \frac{\sqrt{x}-\sqrt{x+h}} {\sqrt{x}\sqrt{x+h}} \)
\( = \lim\limits_{h\to 0} \frac{1}{h} \frac{\sqrt{x}-\sqrt{x+h}}{\sqrt{x}\sqrt{x+h}} \frac{\sqrt{x}+\sqrt{x+h}}{\sqrt{x}+\sqrt{x+h}} \)
\( = \lim\limits_{h\to 0} \frac{1}{h} \frac{x-(x+h)}{\sqrt{x}\sqrt{x+h}} \frac{1}{\sqrt{x}+\sqrt{x+h}} \)
\( = \lim\limits_{h\to 0} \frac{-1}{\sqrt{x}\sqrt{x+h}} \frac{1}{\sqrt{x}+\sqrt{x+h}} = \)
\( = \frac{-1}{x} \frac{1}{2 \sqrt{x}} = \frac{-1}{2x^{3/2}} \)
At \( x=1, y=1, m=f'(1) = \frac{-1}{2} \rightarrow y=\frac{-1}{2}x +\frac{3}{2}\)
At \( x=4, y=\frac{1}{2}, m=f'(4) = \frac{-1}{16} \rightarrow y=\frac{-1}{16}x +\frac{3}{4}\)
Second Quiz
\( f(x) = \frac{x-1}{x+1} \)
\( f'(x)=\lim\limits_{h\to 0} \frac{f(x+h)-f(x)}{h} \)
\( = \lim\limits_{h\to 0} \frac{1}{h} \left( \frac{x+h-1}{x+h+1} -\frac{x-1}{x+1} \right) \)
\( = \lim\limits_{h\to 0} \frac{1}{h} \frac{x^2+xh-x+x+h-1-x^2-xh-x+x+h+1}{(x+h+1)(x+1)} \)
\( = \lim\limits_{h\to 0} \frac{1}{h} \frac{2h}{(x+h+1)(x+1)} \)
\( = \lim\limits_{h\to 0} \frac{2}{(x+h+1)(x+1)} = \frac{2}{(x+1)^2} \)
Set the derivative equal to \( \frac{1}{2} \), then solve for \( x \).
\( f'(x) = \frac{1}{2} \rightarrow x =1 \text{ or } x=-3 \).
Remember to find the \(y-\)coordinates.
\( (1,0), (-3,2) \) are the points on the curve where the slope is \( 1/2 \).
October 10, 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.
P1.
Set \( f(x) = \cos\sqrt{x} - e^x + 2. \) Note that this function is a sum or difference of continuous functions in their respective domains.
Therefore \( f(x) \) is continuous on the interval [0, 2]. We can apply the Intermediate Value Theorem. (Be sure the mention it by name!).
\( f(0) = 1 - 1 + 2 = 2 > 0 \)
\( f(2) \text{ is some negative number because } -e^2 \) is closer to -9.
Therefore, by IVT, the function has at least one zero in the open interval (0, 2), which is the same as saying the equation has at least one solution in the interval (0,2).
P2.
a) \( \lim\limits_{x\to\infty} \frac{\sin^2x}{x^2+1} = 0 \) because the numerator ranges from 0 to 1 and the numerator goes to positive infinity. The ratio therefore goes to zero.
Note 1: do not break this into a quotient of limits. We can only do that when the individual limits exist. In this case, the limit of the numerator is not some finite value. i.e. it does not exist!
An alternate approach here is to use the Squeeze Theorem. Start with the double inequality: \( 0 \leq \sin^2 x\leq 1 \), then multiply by \( \frac{1}{x^2+1} \).
b)
\( \lim\limits_{x\to 0} x^2 \cos\left(\frac{1}{x^2}\right) = 0 \)
Be sure to use the Squeeze Theorem here. Mention it by name and show the work. The example is almost identical to what we did in class.
c)
\( \lim\limits_{x\to (\pi/2)^+} e^{\sec x} = 0 \) because the secant expression goes towards negative infinity as cosine tends towards 0 from the left side.
Overall, the limit goes to zero. If you are still struggling with trigonometric expressions, that's a strong hint that you need to review/relearn the precalculus chapters. We will be seeing A LOT OF trigonometry in this class.
October 7, 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.
P1.
After factoring completely, start with the domain of the function: \( f(x) = \frac{(x-8)(x+2)}{(x-8)(x+1)\sqrt{(x-2)(x+2)}} \).
The domain of the function is \( (-\infty, -2) \cup (2, 8) \cup (8, \infty) \), so we should(!) ignore \( x = -1 \).
The function has an infinite discontinuity (vertical asymptote) at \( x = 2 \) and a removable discontinuity at \( x = 8 \). (FULL CREDIT if you got this far.)
The function approaches 0 as x goes to -2 from the left. As an exercise, you can confirm this by setting up the one-sided limit. The graph has a hole at \( (-2, 0) \), but because the function is not defined on both sides, this point is neither a jump nor a removable discontinuity.
A desmos plot of the graph of the function is below:
P2.
a)
\( \lim\limits_{t\to \pi/2^+} e^{\tan t} = e^{\lim\limits_{t\to \pi/2^+}\tan t} = e^{-\infty} = 0 \)
Feel free to support your reasoning either by showing a graph of \( f(t) = \tan t \) or the unit circle.
b)
\( \lim\limits_{\theta\to 0} \left( \frac{1-\cos\theta} {\theta}\right) = \lim\limits_{\theta\to 0} (-1) \left( \frac{\cos\theta - 1} {\theta}\right) = (-1) * 0 = 0 \)
Optionally, after checking that the limit form is \( \frac{0}{0} \) (use arrows!), multiply both top and bottom by the conjugate of the numerator: \( (1+\cos\theta) \), and then simplify using the fundamental trig identity and the fact that \( \lim\limits_{\theta \to 0} \frac{\sin \theta}{\theta} = 1 \)
c)
Check that the form is 0 over 0 (using arrows), then manipulate the limit as follows:
\( \lim\limits_{x\to 0}\frac{3x - \sin(kx)}{x} =\lim\limits_{x\to 0}\frac{3x}{x} - \lim\limits_{x\to 0}\frac{\sin(kx)}{x} = 3 - k \lim\limits_{x\to 0}\frac{\sin(kx)}{kx} = 3 - k*1 = 3 - k \).
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.