MAT 136, Calculus I

Prof. Swift, Fall 2024

Schedule

Precalculus Review, Differentiation Shortcuts, and Desmos derivative checker.

Tuesday, November 12: Set 22, The Indefinite Integral
Section 5.1: Indefinite Integrals
Section 5.2: Computing Indefinite Integrals

Definition: A function \(F(x)\) is an anti-derivative of the function \(f(x)\) provided that \(F'(x) = f(x)\).

Fact: If \(F(x)\) is an anti-derivative of \(f(x)\) then the most general anti-derivative of \(f(x)\) is \(F(x) + C\), where \(C\) is an arbitrary constant.
Definition: The most general anti-derivative of \(f(x)\) is called the indefinite integral of \(f(x)\), and is denoted \(\displaystyle \int f(x) \, dx = F(x) + C\), where \(F\) is any antiderivative of \(f\) and \(C\) is an arbitrary constant.

Finding anti-derivatives, which we call “doing integrals,” can be tricky. First of all, know the derivative shortcuts. Review these now; the Final Exam is just a month away. We will learn tricks for doing integrals throuout this semester, and continue in Calc 2.
We undo the power rule \(\frac d {dx} x^c = c x^{c-1}\) so often that it is useful to re-phrase it with a shifted exponent. Let \(c = a + 1\) and this becomes
\(\displaystyle \int x^a \, dx = \frac{x^{a+1}}{a+1} + C\), provided \(a \neq -1\). Note that if \(a = -1\) then \(a + 1 = 0\) and we cannot divide by zero. Instead, use the known fact that \(\displaystyle \frac{d}{dx}\ln|x| = \frac 1 x\) to get the integral formula
\(\displaystyle \int x^{-1}\, dx = \int \frac{1}{x} \, dx = \ln|x| + C\).

Some properties of the indefinite integrals:
\(\displaystyle \int c f(x) \, dx = c \int f(x) \, dx, \quad \int f(x) + g(x) \, dx = \int f(x)\, dx + \int g(x)\, dx\). Warning! \(\displaystyle \int f(x) \cdot g(x) \, dx \neq \int f(x) \, dx \cdot \int g(x) \, dx\).
Very Important! There is no product rule for integrals. Calc 2 teaches about integration by parts, which sort of undoes the product rule. But there is no formula for \(\displaystyle \int f(x) \cdot g(x) \, dx\), and there is no formula for \(\displaystyle \int \frac{f(x)}{g(x)} \, dx\).

Here are the solutions to Friday’s quiz. In today’s worksheet on the the Indefinite Integral we find the antiderivatives of those same functions. Here are the scanned solutions.

Friday, November 8: Set 21, Area and the Definite Integral, continued
The Limit Definition of the Definite Integral:
\(\displaystyle \int_a^b f(x) dx = \lim_{n \to \infty} \sum_{i = 1}^{n}f(x_i) \Delta x \), where \(\displaystyle \Delta x = \frac{b-a}{n}\), and \(x_i = a + i \Delta x\).
Note: This definition used the sample point \(x_i^* = x_i\) in the general Riemann Sum. If \(f\) is continuous, then any choice of sample points gives the same limit as \(n \to \infty\).

This Interactive Riemann Sum Applet shows the rectangles.

Some properties of the definite integral: \(\displaystyle \int_a^b c f(x) \, dx = c \int_a^b f(x) \, dx, \quad \int_a^b f(x) + g(x) \, dx = \int_a^b f(x)\, dx + \int_a^b g(x)\, dx, \quad \int_a^b f(x) \, dx + \int_b^c f(x) \, dx = \int_a^c f(x) \, dx\)

In-class worksheet on The Definite Integral, Part 2, with the scanned solutions.
Here is a video showing how Gauss discovered \(\displaystyle \sum_{i=1}^n i = \frac{n(n+1)}{2}.\) For your last problem on the WeBWorK set you will need the fact that \(\displaystyle \sum_{i=1}^n i^2 = \frac{n(n+1)(2n+1)}{6}.\)

Wednesday, November 6: Set 21, Area and the Definite Integral
Section 5.5: The Area Problem
Section 5.6: Definition of the Definite Integral.
The Geometric Definition of the Definite Integral:
\(\displaystyle \int_a^b f(x) dx \) is the signed area between \(y = 0 \) and \(y = f(x)\) with \(a \leq x \leq b\). (The integral is the “area under the curve.”)
A Riemann Sum to approximate the definite integral:
\(\displaystyle \sum_{i = 1}^{n}f(x_i^*) \Delta x \), where \(\displaystyle \Delta x = \frac{b-a}{n}\), \(x_i = a + i \Delta x\), and for each \(i\), the sample point \(x_i^*\) is any point in the sub-interval\([x_{i-1}, x_i]\).
We consider three cases for the Riemann sum:
The Left sum \(\displaystyle L_n = \sum_{i = 0}^{n-1}f(x_i) \Delta x \), the Right sum \(\displaystyle R_n = \sum_{i = 1}^{n}f(x_i) \Delta x \), and the Midpoint sum \(\displaystyle M_n = \sum_{i = 1}^{n}f(\overline{x_i}) \Delta x \), where \(\displaystyle \overline{x_i} = \frac{x_{i-1}+ x_i}{2}\) is the midpoint of the sub-interval.
In-class worksheet on The Definite Integral, with the scanned solutions.

Friday, November 1: Set 20, Optimization and Newton's Method
Section 4.13: Newton's Method.
Newton's Method: To find an approximate solution to \(f(x) = 0\), start with a guess \(x_0\) and compute \(x_1\), \(x_2\), \(x_3\), etc. using \(\displaystyle x_{n+1} = x_n - \frac{f(x_n)}{f'(x_n)}\). Stop when \(x_{n+1} \approx x_n\) to the desired accuracy, since it follows that \(f(x_n) \approx 0\).
Here is a YouTube video of me showing how to do Mewton's Method with a Spreadsheet.

Wednesday, October 30: Set 20, Optimization
Section 4.8: Optimization, and Section 4.9: More Optimization.
The key is setting up a function of one variable, \(Q = f(x)\), whose output \(Q\) is the thing you want to optimize. Here are some steps:
(1) Understand the problem (Read it carefully, and frequently)
(2) Draw a diagram. Identify the given fixed quantities, and the variable quantities.
(3) Introduce notation. The word "Let" is super-important.
(4) Write the quantity you want to optimize (we will call it \(Q\)) in terms of other quantites.
(5) Usually \(Q\) will depend on more than one quantity. Use the constraints to eliminate all but one variable quantity to get \(Q = f(x)\). Write down the domain of \(f\).
(6) Use calculus to find the global max (or min) value of \(f\), or the input \(c\) at which the global extremum occurs. Read the question again and answer it.

Tuesday, October 29: More on set 19, L'Hospital's Rule
The indeterminate forms are
\( \frac 00,~ \frac{\infty}{\infty},~ 0\cdot\infty,~ \infty - \infty,~ 0^0,~ 1^\infty, \text{ and } \infty^0 .\)
L'Hospital's rule can only be used for limits of type \(\frac 00\) and of type \(\frac{\infty}{\infty}\).
In-class worksheet on Indeterminate Forms, with the scanned solutions.

Monday, October 28: Set 19, Section 4.10: L'Hospital's Rule
L'Hospital's Rule, also written L'Hôpital's Rule, says that if \(\displaystyle \lim_{x \to a} f(x) = \lim_{x \to a} g(x) = 0\), or \(\displaystyle \lim_{x \to a} f(x) = \pm \infty\) and \(\displaystyle \lim_{x \to a} g(x) = \pm \infty\), then
\(\displaystyle \lim_{x \to a} \frac{f(x)}{g(x)} = \lim_{x \to a} \frac{f'(x)}{g'(x)}\)
L'Hospital's Rule can be applied to \(\displaystyle \lim_{x \to \infty} \frac{f(x)}{g(x)}\) and \(\displaystyle \lim_{x \to -\infty} \frac{f(x)}{g(x)}\), provided they are indeterminate forms of type \(\frac 00\) or type \(\frac{\infty}{\infty}\).
In-class worksheet on L'Hospital's rule, with the scanned solutions. Here is a Desmos Graph.

Friday, October 25: Set 18, Section 4.6 : The Shape Of A Graph, Part II

Definition: A function \(f\) is concave up on an interval \(I\) provided that every secant line with \(x\)-values in \(I\) is above the graph of \(f\).
Similarly, \(f\) is concave down if every secant line is below the graph.

Examples: Here is the graph of y = exp(x/2), which is concave up on \((-\infty, \infty)\).
This function \(f\) is concave up on \((-\infty, \infty)\) even though \(f\) is not differentiable at 0.
This function is concave up on \((-\infty, 0]\) and also concave up on \([0, \infty)\), but is not concave up on \((-1,1)\).

We will usually consider functions that are differentiable everywhere so the graph has no corners.

Theorem: A function \(f\) is concave up on an interval \(I\) if \(f'\) is increasing on \(I\).

Theorem: A function \(f\) is concave up on an interval \(I\) if \(f''\) is positive on \(I\).
Similar theorems holds for concave down.

Let \(I\) be an interval in what follows.
If \(f' > 0\) on \(I\), then \(f\) is increasing on \(I\).
If \(f' < 0\) on \(I\), then \(f\) is decreasing on \(I\).
If \(f'' > 0\) on \(I\), then \(f'\) is increasing on \(I\), and \(f\) is concave up on \(I\).
If \(f'' < 0\) on \(I\), then \(f'\) is decreasing on \(I\), and \(f\) is concave down on \(I\).

There are two more cases:
If \(f' = 0\) on \(I\), then \(f\) is constant on \(I\).
If \(f'' = 0\) on \(I\), then \(f'\) is constant on \(I\), and \(f\) is straight on \(I\).

Definition: An inflection point of \(f\) is a point on the graph of \(f\), namely \((c, f(c))\), where the concavity of \(f\) changes.

Theorem: If \(f''(c) = 0\), and \(f''\) changes sign at \(c\), then \((c, f(c))\) is an inflection point of \(f\).

Theorem: If \(f'\) has a local extremum at \(c\), then \((c, f(c))\) is an inflection point on the graph of \(f\).

Theorem: (The Second Derivative Test) If \(f'(c) = 0\) and \(f''(c)>0\), then \(f\) has a local minimum at \(c\).
Similarly, (\(f'(c) = 0\) and \(f''(c)< 0 ) \implies \) \(f\) has a local maximum at \(c\).
Note that if \(f'(c) =0 \) and \( f''(c) = 0\), then the second derivative test says absolutely nothing.

In-class worksheet on the Shape of Graphs. This is worth 5 class points and will be collected at the beginning of class on Monday. Here are the scanned solutions. Here is a desmos graphs of \(f\), \(f'\) and \(f''\) for \(f(x) = x^3 -3x^2 + 1\), which is the subject of that worksheet.

Tuesday and Wednesday, October 22 and 23: Set 17, Critical Points and Local Extrema, and the Mean Value Theorem.
Section 4.3: Minimum and Maximum Values,
Section 4.5: The Shape of a Graph, Part I.
Section 4.7: Mean Value Theorem. (I will only cover this briefly in class. Read Paul's notes.)

Definition: A function \(f\) is increasing on an interval \(I\) provided that \(f(x_1) < f(x_2)\) for all \(x_1 < x_2\) in \(I\).
A similar definition holds for decreasing.
An interval has the form \( I = (a, b)\) or \( I = [a, b)\) or \( I = (-\infty, b)\) or \( I = [a, \infty)\), etc. An interval has no holes, and it cannot be a single point.

Examples:
\(f(x) = x^2\) is increasing on \([0, \infty)\) and decreasing on \( (-\infty, 0] \). Note that both intervals include the point \(x = 0\).
Similarly, \(g(x) = x^3\) is increasing on \( (-\infty, \infty)\).

Theorem: If \(f'(x) > 0\) for all \(x\) in an interval \(I\), then \(f\) is increasing on \(I\).
A similar theorem holds: \(f' < 0\) on \(I\) implies that \(f\) is decreasing on \(I\).

Examples:
\(f(x) = x^3 + x\) has the derivative \(f'(x) = 3 x^2 + 1\), which is positive on \((-\infty, \infty)\), so \(f\) is increasing on \((-\infty, \infty)\).
This theorem implies that \(f(x) = x^2\), for which \(f'(x) = 2x\), is increasing on \((0, \infty)\) and decreasing on \((-\infty, 0)\). This is the truth, but not the whole truth.
Similarly, the theorem says that \(g(x) = x^3\), for which \(g'(x) = 3 x^2\), is increasing on \((0, \infty)\) and also increasing on \((-\infty, 0)\)

We can get a stronger result:
Theorem: If \(f'(x) \geq 0\) for all \(x\) in an interval \(I\), and \(f'(x) = 0\) only at a finite number of points (possibly no points), then \(f\) is increasing on \(I\).
A similar theorem holds for \(f'(x) \leq 0\) and \(f\) decreasing.

Examples: We can see from the definition that \(f(x) = x^2\) is increasing on \( [0, \infty) \), and the stronger theorem proves this, since \(f'(x) = 2x \geq 0\) on \( [0, \infty) \), and only \(x = 0\) satisfies \(f'(x) = 0\).
Similarly, the stronger theorem says that \(g(x) = x^3\) is increasing on \( (-\infty, \infty)\), since \(g'(x) = 3 x^2 \geq 0\) for all \(x\), and the only solution to \(g'(x) = 0\) is \(x = 0\).

In-class worksheet for Tuesday, Sept. 22: Find the largest interval(s) on which the following functions are increasing.
\(f(x) = \frac 1 3 x^3 - \frac 1 2 x^2 -2 x\) and \(g(x) = \frac 1 3 x^3 - x^2 + x\). Here are the scanned solutions.
Here are desmos graphs for the functions f and g.

Wednesday
Definition: We say a function \(f\) has a local maximum at \(c\) if \(f(c) > f(x)\) for all \(x\) sufficiently close to, but not equal to, \(c\)
A similar definition holds for a local minimum of \(f\) at \(c\).

Theorem: (The First Derivative Test) Suppose that \(f\) is continuous at \(c\).
1. If \(f'(x) > 0\) for \(x < c \) and \(f'(x) < 0\) for \(c < x\), then \(f\) has a local maximum at \(c\).
2. If \(f'(x) < 0\) for \(x < c\) and \(f'(x) > 0\) for \(c < x \), then \(f\) has a local minimum at \(c\).
3. If \(f'(x) \) has the same sign for \(x < c\) and \(c < x\), then \(f\) does not have a local extremum at \(c\).
Note: Those inequalities about the sign of \(f'(x)\) only need to hold for \(x\) close to \(c\).

Theorem: If \(f\) has a local extremum at \(c\), then \(c\) is a critical point of \(f\).
The converse is not true. If \(c\) is a critical point of \(f\), then \(f\) might or might not have a local extremum at \(c\).

In-class worksheet on The First Derivative Test to Classify Critical Points. Here are the scanned solutions.
Here is a desmos graph of f and f-prime.
Here is a desmos graph of g and g-prime.

Friday and Monday, October 18 and 21: Section 4.2: Critical Points and Section 4.4: Finding Global Extrema (WeBWorK set 16).
Definition: A critical point of a function \(f\) is a number \(c\) such
(1) \(f'(c)=0\) or \(f'(c)\) is undefined, and
(2) \(f(c)\) is defined. (That is, \(c\) is a point in the domain of \(f\).)
Definition of Global Extrema, and a theorem about global extrema of a continuous function on a closed interval.
In-class worksheet on Critical Points and Global Extrema, and the scanned solutions.

Wednesdau, October 16: Section 4.11: Local Linearization and the Tangent Line Approximation. WeBWorK set 15.
The local linearization of \(f\) at \(a\) is \(\ell_a(x) = f(a) + f'(a)(x-a)\).
Sometimes the notation \(L_a(x)\) or \(L(x)\) is used instead of \(\ell_a(x)\).
(1) \(y = \ell_a(x)\) is an equation of the tangent line to the graph of \(f\) at \(a\).
(2) \(f(x) \approx \ell_a(x)\) for \(x \approx a\) is the tangent line approximation.
For a linear function, the tangent line approximation is exact.
Here is one of Richard Feynman's many amusing stories: Feynman vs. the Abacus.
In-class worksheet on Local Linearization. Here are the scanned solutions.

Friday, Oct. 11: Logarithmic Differentiation, and Differentiation Review
Section 3.13 : Logarithmic Differentiation.
Let \(y = f(x)\), take the natural logarithm of both sides, and then do implicity differentiation to find \(\frac{dy}{dx} = f'(x)\).

Here are the scanned solutions to Friday's quiz

Wednesday, Oct. 9: more on WeBWorK set 14
Section 3.6 : Derivatives of Exponential and Logarithmic Functions,
Section 3.7 : Derivatives Of Inverse Trig Functions,
Today's facts, derived using the chain rule or implicit differentiation.

\(\displaystyle \frac d {dx} a^x = \ln(a) a^x\) for constant \(a > 0\),     \(\displaystyle\frac d {dx} \ln(x) = \frac 1 x \) with domain \(x > 0\),     \(\displaystyle \frac d {dx} \ln|x| = \frac 1 x \) with domain \(x \neq 0\).

\(\displaystyle \frac d {dx} \arcsin(x)= \frac {1}{\sqrt{1-x^2}}\),     \(\displaystyle \frac d {dx} \arctan(x)= \frac {1}{1+x^2}\).

Note: WeBWorK frequently uses \(\sin^{-1}(x)\) as a notation for \(\arcsin(x)\), and \(\tan^{-1}(x)\) as a notation for \( \arctan(x)\).
Desmos graphs for arcsine and arctangent
Worksheet on derivatives of inverse trig functions, and implicit differentiation, with the scanned solutions. Here is a desmos graph of the tangent line to the curve.

Tuesday, Oct. 8: Start WebWorK set 14
Section 3.10 : Implicit Differentiation.

Implicit differentiation: Start with an equation involving \(x\) and \(y\). (So \(y\) is an implicit function of \(x\).) Use the chain rule to differentiate both sides of the equation with respect to \(x\), treating \(y\) as an unknown function of \(x\). Then, solve for \(\frac{dy}{dx}\).

Friday, Oct. 4 and Monday, Oct. 7: WeBWorK set 13 Section 3.9 : Chain Rule
The chain rule can be stated in two equivalent ways:
If \(y = f(u)\) and \(u = g(x)\), then \(\frac{dy}{dx} = \frac{dy}{du} \frac{du}{dx}\)
\(\frac{d}{dx} f(g(x)) = f'(g(x))\cdot g'(x) \)

Here are the solutions to the in-class worksheet from Oct. 4.

Monday, Oct. 7: More on Set 13

Here is the verbal Chain Rule.
Here is a page with a derivative table from a book published in 1933. Note how the chain rule is combined with each derivative fact.
Today's on-board examples, with the scanned solutions
Here are desmos checks of problem 1, problem 2, and problem 3.

Tuesday and Wednesday, Oct. 1 and 2: WeBWorK set 12
Set 12 covers Section 3.12 : Higher Order Derivatives and Section 3.5 : Derivatives Of Trig Functions.

Today's notation:
\(f''(x) = \frac{d}{dx}[f'(x)] = \frac{d}{dx} \left [ \frac{d}{dx} [f(x)] \right ] = \frac{d^2}{dx^2} [f(x)]\).

Today's facts:
\(\frac d {dx} \sin(x) = \cos(x)\), \(\frac d {dx} \cos(x) = -\sin(x)\), \(\frac d {dx} \tan(x) = \frac 1 {\cos^2(x)}\).

Everything you need to know about computing derivatives in this class are summarized in the page Differentiation Shortcuts.
Note that you can use trig identities to avoid memorizing the derivatives of \(\csc(x)\), \(\sec(x)\) and \(\cot(x)\).
Derivative Shortcuts 3, with the scanned solutions

Wednesday: Desmos graph for the Numerical derivative of h(x) = sin(x)
Here is a multi-purpose desmos graph showing the numerical derivative of any function. You can use this to check your calculation of the derivative.

Here is today's worksheet, with the scanned solutions

Friday, Sept. 27, and Monday, Sept. 30: WeBWorK set 11
Set 11 covers Section 3.2 : Interpretation of the Derivative. In this webwork we take a break from differentiation shortcuts.

The derivative \(f'(x) = \frac{dy}{dx}\) is the rate of change of \(f\), and the slope of the tangent line to \(y = f(x)\). The units of \(\displaystyle \frac{dy}{dx} = \lim_{\Delta x \to 0} \frac{\Delta y}{\Delta x} \) are the units of \(y\) divided by the units of \(x\).

Here are scanned solutions to Friday's Quiz.
Worksheet on the Interpretation of the derivative

Wednesday, Sept. 25: More on WeBWork set 10
Rule:
\( \displaystyle \frac{d}{dx} \left [ \frac {f(x)}{ g(x)} \right ] = \frac{f'(x)\cdot g(x) - f(x) \cdot g'(x) }{g(x)^2} \quad\) The quotient rule

Here are the verbal Product and Quotient Rules.

Here is a proof of the Product Rule and the Quotient Rule.

Worksheet on Product and Quotient Rules, with the scanned solutions

Tuesday, Sept. 24: More on WeBWork set 9, start set 10
Set 10 covers Section 3.4 : Product And Quotient Rule

Rule:
\(\frac{d}{dx} [f(x)\cdot g(x)] =f'(x)\cdot g(x) + f(x) \cdot g'(x) \quad\) The product rule. Don't make this mistake.
Fact:
\(\frac{d}{dx} e^x = e^x \quad \) The derivative of the natural exponential function is itself!
Worksheet on Derivative Shortcuts 2, with the scanned solutions.

Monday, Sept. 23: The Derivative as a Function, WeBWork set 9
We start to learn how to differentiate any nice function, using Section 3.3: Differentiation Formulas.
Rules:
\(\frac{d}{dx} [c f(x)] = c f'(x) \quad\) The constant multiple rule
\(\frac{d}{dx} [f(x) \pm g(x)] = f'(x) \pm g'(x) \quad \) The sum and difference rules
Facts:
\(\frac{d}{dx} [c] = 0 \quad \) The derivative of a constant function is the 0 function
\(\frac{d}{dx} [x^c] = c x^{c-1} \quad \) The Power Rule
Worksheet on Derivative Shortcuts 1, with the scanned solutions

Monday, Sept 16 and Tuesday, Sept 17: The definition of the derivative, WeBWork set 8
Section 3.1: The definition of the Derivative.
Note: A sample exam for Friday's Exam 1 is on Canvas in the "files" section.

Here are the solutions to Friday's quiz.

Definition: The derivative of the function \(f\) at \(a\) is the slope of the tangent line to \(y = f(x)\) at the point (\(a, f(a))\).
The notation for the derivative is \(f'(a)\), pronounced "\(f\) prime of \(a\)", and it can be computed as
\(\displaystyle f'(a) = \lim_{x \to a} \frac{f(x) - f(a)}{x-a}\).

Replace \(x\) with \(a + h\), so \(\Delta x = x -a = (a+h) - a = h\), and this becomes
\(\displaystyle f'(a) = \lim_{h \to 0} \frac{f(a+h) - f(a)}{h}\). This is illustrated in this Desmos graph.

Since \(a\) can be any point in the domain of \(f\), we can replace \(a\) by \(x\) and get our most useful way to compute the derivative.
\(\displaystyle f'(x) = \lim_{h \to 0} \frac{f(x+h) - f(x)}{h}\).
Given a function \(f\), this defines a new function, \(f'\).

Today's in-class group work: Find an equation of the tangent line to the graph of \(f(x) = x^2 - 4x + 7\) at the point \( (3, f(3)) \).
Hints: Compute \(f'(3)\) using the definition \(f'(3) = \displaystyle \lim_{h \to 0} \frac{f(3+h)-f(3)}h \), showing the steps as carefully as in the solutions to the quiz. Then use the point-slope formula for an equation of the tangent line.
Solution: Since \(f(3) = 4\), the point-slope form of the tangent line is \(y = m(x-3) + 4\). The slope of the tangent line is
\(m = f'(3) = \displaystyle \lim_{h \to 0} \frac{f(3+h)-f(3)}h = \lim_{h \to 0} \frac{(3+h)^2 - 4(3+h) + 7 \ -4}h = \lim_{h \to 0} \frac{9 + 2\cdot 3 h + h^2 - 12 - 4h + 7 \ -4}h = \lim_{h \to 0} \frac{6 h + h^2 - 4h}h = \lim_{h \to 0} \frac{h(2 + h)}h = \lim_{h \to 0} (2 + h) = 2 + 0 = 2. \)
To summarize, the slope of the tangent line is \(f'(3) = 2\). Thus an equation of the tangent line is \(y = 2(x-3)+4\). Here is a graph of both \(f\) and the tangent line at \(x = 3\).
This point-slope form of the tangent line is perferred, in calculus, to \(y = 2x - 2\).

Friday, September 13: Finish More Limets, WeBWorK set 7.
The Intermediate Value Theorem (IVT): From Section 2.9: Continuity
Suppose \(f(x)\) is continuous on \([a,b]\), and \(M\) is a number strictly between \(f(a)\) and \(f(b)\). Then there exists a number \(c\) such that
\(a < c < b\), and \(f(c) = M\).

The Squeeze Theorem: From Section 2.5: Computing Limits
Assume \(a < b < c\) and \(f_\ell(x) \leq f(x) \leq f_u(x)\) for all \(x \in (a,b)\cup( b,c)\), and \(\displaystyle \lim_{x \to b} f_\ell(x) = \lim_{x \to b} f_u(x) = L\). Then \(\displaystyle \lim_{x \to b} f(x) = L\).
The squeeze theorem can be used to compute limits where the algebraic manipulation method fails. A famous example is the proof that \(\displaystyle \lim_{x\to 0} \frac{\sin(x)}{x} = 1\). Please watch this Khan Academy video:
limit of sin(x) over x as x approaches 0.

Wednesday, September 11: Finish set 5 and start More Limits, WeBWorK set 7

Section 2.7 : Limits At Infinity, Part I
Section 2.8 : Limits At Infinity, Part II
Worksheet 9, Limits at infinity, with the scanned solutions

Tuesday, September 10: Continuity and Algebraic Limits, WeBWorK set 6
The pencil and paper technique for computing limits.
Suppose \(f(a)\) is not defined and you are asked to compute \(\displaystyle \lim_{x \to a} f(x) \).
Do algebraic manipulation on the function \(f(x)\) to "fill in the hole" in the function. Call this new function \(\tilde f\). It is a different function because it has a different domain. Hopefully the function \(\tilde f\) is continuous at \(a\), so the original limit is \(\tilde f(a)\).
Example: Compute \(\displaystyle \lim_{x \to 1} \frac{x^2 - 1}{x-1}\)

Here is a technical way to describe the technique:
If \(f(x) = \tilde f(x)\) for all \(x \neq a\), and \(\tilde f\) is continuous at \(a\), then \(\displaystyle \lim_{x \to a} f(x) = \displaystyle \lim_{x \to a} \tilde f(x) = \tilde f(a) \).
Here is the desmos graph of one version of set 6, problem 12. You can play around with the sliders for \(a\) and \(b\) and try to make the function \(f\) continuous at both 2 and 3, but that is too hard!
The key is to get 2 equations in the 2 unknowns \(a\) and \(b\) by imposing the condition that \(f\) is continuous at 2 and 3. Here is the solution to my version of this problem. The solution to the system of 2 equations is \(a = -1\), \(b = -4\). You can go to that desmos graph and use the sliders to get a continuous function.

Worksheet 7, along with the scanned solutions.

Monday, 9-9: Questions on set 4 and start
Continuity and Algebraic Limits, WeBWorK set 6
These sections in Paul's notes will be helpful:
Section 2.4: Limit Properties
Section 2.5: Computing Limits
Section 2.9: Continuity

Friday, September 6: Graphical Limits, WeBWorK set 5
These sections of Paul's notes will be helpful:
Section 2.2: The Limit
Section 2.3: One-Sided Limits
Section 2.6: Infinite Limits
Rules for sketching graphs: An open circle stands for a point on the graph. A closed circle stands for a point not on the graph.
\(\displaystyle \lim_{x \to 2} f(x) = 3\) means that \(f(x) \approx 3\) for all \(x \approx 2\), except possibly \(x = 2\).
An alternative notation is \(f(x) \to 3\) as \(x \to 2\).
Note: The value of \(f(2)\), has nothing to do with the limit as \(x\) approaches 2. Frequently, we are asked to determine \(\displaystyle \lim_{x \to 2} f(x) \) in the cases where \(f(2)\) is not defined.
We will have group work, worth 5 class points, during the last 15-20 minutes of class.
Here is the in-class Worksheet 6. Here are the scanned solutions to the quiz.

Wednesday, September 4: Rates of Change, WebWorK set 4
(1) How to numerically estimating the slope of the tangent line to a graph, which is the average rate of change of the fuction.
(2) How to graphically estimate the slope of the tangent line to a graph.
We also work on Problem 4 in set 4.

Tuesday, September 3: Rates of Change, WebWorK set 4:
We finally start Calculus!
Here is the introduction to Paul's Notes Chapter 2. Set 4 is based on Paul's Section 2.1 : Tangent Lines And Rates Of Change, but Set 4 has more of an emphasis on velocity as the rate of change of position.
\(\displaystyle m_{PQ} =\frac{\Delta y}{\Delta x}\) is the slope of the secant line through points \(P\) and \(Q\) on the graph of a function. Furthermore, \(m_{PQ}\) is the average rate of change of the function. The slope of the tangent line at \(P\) is the "limit" of \(m_{PQ}\) as \(Q\) approaches \(P\). This slope is also the instantaneous rate of change of the function.

Note that \(\frac{\Delta y}{\Delta x}\) is undefined when \(\Delta x = 0\). Calculus is the story of how we handle this difficulty. We will define "limit" in the future. In Set 4 we mainly compute \(\frac{\Delta y}{\Delta x}\) when \(\Delta x\), often called \(h\), is small but nonzero, and we sometimes guess the limit of \(\frac{\Delta y}{\Delta x}\) as \(h\) approaches 0.

Friday, August 30: Composition of Functions and Inverse Functions, WeBWorK set 3:
Definition of open and closed intervals: (Needed in webwork set 3, problem 1)
\( (a, b) = \{x \in \mathbb{R} \mid a < x < b\}\). The open interval, written with parentheses, does not include endpoints.
\( [a, b] = \{x \in \mathbb{R} \mid a \leq x \leq b\}\). The closed interval, written with square brackets, does include endpoints.
\( (a, b] = \{x \in \mathbb{R} \mid a < x \leq b\}\). This interval does not include \(a\), but it does include \(b\). This interval is neither open nor closed.
Worksheet 4. Here are the scanned solutions.

Wednesday, August 28: Trig functions and function notation, WeBWorK set 2:
Worksheet 3 on Linear, Polynomial, and Exponential functions.
Here are the scanned solutions to the first 2 problems. I gave a hint for problem 14 on the WeBWorK in class.
Here is a hint for problem 15. Try these problems, and then look at the solutions.

Tuesday, August 27: Linear and Exponential functions, WeBWorK set 1:
Note: We did not do worksheet 2, because of technology problem.
In class worksheet 2 on the Point-Slope Form for Linear Functions. scanned solutions.

Monday, August 26: Introduction:
Required knowledge from Precalculus. Here is the Precalculus Chapter in Paul's Notes.
Today's ice-breaker worksheet is on paper.