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MTH191 Calculus课程简介

This is a first-semester single variable college calculus course. Students learn topics including limits, L’Hôpital’s Rule, continuity, transcendental functions, derivatives, antiderivatives, applications of derivatives, including optimization, maximum and minimum problems, business, economics, sports, engineering, physics, Newton’s method, and applications of integration. A graphing calculator is required for this course. See the time schedule for the current brand and model.

PREREQUISITES 

As a language model, I don’t have access to specific time schedules or brand and model of graphing calculators. However, I can tell you that the topics you mentioned are all fundamental concepts in a first-semester single variable college calculus course.

In this course, students will begin by studying limits and continuity, which are crucial concepts for understanding the behavior of functions. They will also learn about derivatives, which describe the instantaneous rate of change of a function, and antiderivatives, which are used to find the area under a curve.

Students will learn how to apply these concepts to a variety of real-world situations, including optimization problems in business and economics, motion problems in physics, and engineering problems involving rates of change.

To aid in their understanding of these concepts, students are typically required to use a graphing calculator, which can help them visualize functions and understand their behavior. Different schools or instructors may require different brands and models of calculators, so it’s important to check the specific requirements for your course.

MTH191 Calculus HELP（EXAM HELP， ONLINE TUTOR）

1. We have $f(-a)=-\frac{1}{a}$, $f\left(a^{-1}\right)=a^2+2a$, $f(\sqrt{a})=\frac{1}{\sqrt{a}}$, and $f\left(a^2\right)=\frac{1}{a^2}$.
2. We have $f(a+h)=(a+h)^2+2(a+h)=a^2+2a+2ah+h^2+2h$, so $f(a+h)-f(a)=(a+h)^2+2(a+h)-(a^2+2a)=2ah+h^2+2h$.
3. The function $f(x)=\frac{2}{3-x}$ is undefined when the denominator $3-x$ equals zero, i.e., when $x=3$. Therefore, the largest domain of real numbers where $f(x)$ determines a real valued function is $(-\infty, 3) \cup (3, \infty)$.
4. The function $f(x)=\sqrt{\frac{x+1}{x-1}}$ is defined only when the radicand $\frac{x+1}{x-1}$ is nonnegative, i.e., when $x>1$. Therefore, the largest domain of real numbers where $f(x)$ determines a real valued function is $(1,\infty)$.
5. Let $y$ be the daily production of each well, then the total daily production of the oil field after drilling $x$ new wells is $4000+20y-5xy=4000+20\left( \frac{4000}{20-x} \right)-5x\left( \frac{4000}{20-x} \right)$. Simplifying, we get $f(x)=8000\frac{x}{20-x}$.
6. If each square has side $x$, then the dimensions of the base of the box are $50-2x$ by $50-2x$, and the height of the box is $x$. Therefore, the volume of the box is $f(x)=x(50-2x)^2$.

Suggested Problem:

36. Let $s$ be the side length of the square and $P$ be the perimeter. Then $P=4s$, so $s=\frac{P}{4}$. The area of the square is $A=s^2=\left(\frac{P}{4}\right)^2$, so $A=\frac{P^2}{16}$. Therefore, the area of the square is a function of its perimeter given by $f(P)=\frac{P^2}{16}$.

5. Write an equation of the line $L$ described and sketch its graph: $L$ passes through $(2,-3)$ and $(5,3)$.

To find the equation of the line $L$, we first need to find its slope. The slope of a line passing through two points $(x_1, y_1)$ and $(x_2, y_2)$ is given by:

$$m=\frac{y_2-y_1}{x_2-x_1}$$

Substituting the given points, we have:

$$m=\frac{3-(-3)}{5-2}=\frac{6}{3}=2$$

Now that we know the slope of the line, we can use the point-slope form of the equation of a line to find its equation. The point-slope form of the equation of a line with slope $m$ passing through the point $(x_1, y_1)$ is given by:

$$y-y_1=m(x-x_1)$$

Substituting the slope and one of the points, we have:

$$y-(-3)=2(x-2)$$

Simplifying and solving for $y$, we get:

$$y=2x-7$$

To sketch the graph of $L$, we can plot the two given points and connect them with a straight line:

39. Sketch the graph of $f(x)=\frac{1}{(x-1)^2}$.

To sketch the graph of $f(x)$, we can start by analyzing its behavior as $x$ approaches the values of the vertical asymptotes, which occur when the denominator of $f(x)$ equals zero:

$$(x-1)^2=0 \implies x=1$$

So, $x=1$ is a vertical asymptote of the graph of $f(x)$. As $x$ approaches 1 from the left, $f(x)$ goes to infinity, and as $x$ approaches 1 from the right, $f(x)$ goes to infinity with a different sign.

Next, we can analyze the behavior of $f(x)$ as $x$ goes to positive or negative infinity. As $x$ gets very large in either direction, the term $(x-1)^2$ dominates the denominator of $f(x)$, and we can approximate $f(x)$ as:

$$f(x) \approx \frac{1}{x^2}$$

So, as $x$ goes to infinity, $f(x)$ approaches zero. Similarly, as $x$ goes to negative infinity, $f(x)$ approaches zero.

Using this information, we can sketch the graph of $f(x)$ as follows:

• Draw a vertical asymptote at $x=1$.
• Draw a horizontal asymptote at $y=0$.
• Plot a point at $(0,1)$, which is the value of $f(x)$ when $x=0$.
• As $x$ approaches 1 from the left, $f(x)$ goes to positive infinity.
• As $x$ approaches 1 from the right, $f(x)$ goes to negative infinity.
• As $x$ goes to positive or negative infinity, $f(x)$ approaches zero.

The resulting graph is shown below:

65. Use the method of completing the square to graph $y=96 t-16 t^2$ and find the maximum value.

We can write the given equation in vertex form by completing the

Textbooks

• An Introduction to Stochastic Modeling, Fourth Edition by Pinsky and Karlin (freely
available through the university library here)
• Essentials of Stochastic Processes, Third Edition by Durrett (freely available through
the university library here)
To reiterate, the textbooks are freely available through the university library. Note that
you must be connected to the university Wi-Fi or VPN to access the ebooks from the library
links. Furthermore, the library links take some time to populate, so do not be alarmed if
the webpage looks bare for a few seconds.

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