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这是美国康涅狄格大学微积分课程的代写成功案例。
Math1131课程简介
Course communication: Much of the communication for this course will occur online, via HuskyCT, email, and/or this page. Please monitor this page, course HuskyCT announcements, and your UConn email regularly for updates related to MATH 1131Q. However, you are also expected to be in class every day to receive important communications there, and to contact a classmate, or your instructor or TA, if you miss class to make sure you have not missed important course information.
Prerequisites
Time Zone: All times for due dates, online and in-person classes, office hours, Supplemental Instruction, and Q Center hours in this course are given in US Eastern Time.
Description: This course is an introduction to differential and integral calculus, which is the mathematical language used in any science concerned with dynamically changing quantities. The main topics it covers are limits, derivatives, integrals, the Fundamental Theorem of Calculus, and some basic applications of these ideas.
Prerequisites: A qualifying score of 22 on the mathematics placement exam (MPE). Students who fail to achieve this minimum score are required to spend time on the preparatory and learning modules before re-taking the MPE, or to register for a lower level Mathematics course. Not open for credit to students who have passed MATH 1132Q or 1152Q.
Math1131 Calculus HELP（EXAM HELP， ONLINE TUTOR）
Find the domain and the range of the function $f$ which is defined by
$$
f(x)=\frac{2-3 x}{7-2 x} .
$$
The domain consists of all points $x \neq 7 / 2$. To find the range, we note that
$$
\begin{aligned}
y=\frac{2-3 x}{7-2 x} & \Longleftrightarrow 7 y-2 x y=2-3 x \quad \Longleftrightarrow \quad 3 x-2 x y=2-7 y \
& \Longleftrightarrow x(3-2 y)=2-7 y \quad \Longleftrightarrow \quad x=\frac{2-7 y}{3-2 y} .
\end{aligned}
$$
The rightmost formula determines the value of $x$ that satisfies $y=f(x)$. Since the formula makes sense for any number $y \neq 3 / 2$, the range consists of all numbers $y \neq 3 / 2$.
Show that the function $f:(0,1) \rightarrow(0,2)$ is bijective in the case that
$$
f(x)=\frac{4 x}{3-x} .
$$
To show that the given function is injective, we note that
$$
\begin{aligned}
\frac{4 x_1}{3-x_1}=\frac{4 x_2}{3-x_2} & \Longrightarrow \quad 12 x_1-4 x_1 x_2=12 x_2-4 x_1 x_2 \
& \Longrightarrow \quad 12 x_1=12 x_2 \quad \Longrightarrow \quad x_1=x_2 .
\end{aligned}
$$
To show that the given function is surjective, we note that
$$
y=\frac{4 x}{3-x} \quad \Longleftrightarrow 3 y-x y=4 x \quad \Longleftrightarrow \quad 3 y=x(y+4) \quad \Longleftrightarrow \quad x=\frac{3 y}{y+4} .
$$
The rightmost equation determines the value of $x$ such that $y=f(x)$ and we need to check that $00$ and also
$$
1-x=1-\frac{3 y}{y+4}=\frac{y+4-3 y}{y+4}=\frac{4-2 y}{y+4}=\frac{2(2-y)}{y+4}>0
$$
so $00$ and also
$$
2-y=2-\frac{4 x}{3-x}=\frac{6-2 x-4 x}{3-x}=\frac{6(1-x)}{3-x}>0 \quad \Longrightarrow \quad 0<y<2
$$
Find the domain and the range of the function $f$ which is defined by
$$
f(x)=\sqrt{4-\sqrt{x}} .
$$
The domain consists of all numbers $x$ with $x \geq 0$ and $4-\sqrt{x} \geq 0$. This gives $\sqrt{x} \leq 4$ and also $x \leq 16$, so the domain is $[0,16]$. To find the range, we note that
$$
y=\sqrt{4-\sqrt{x}} \Longrightarrow y^2=4-\sqrt{x} \quad \Longrightarrow \quad \sqrt{x}=4-y^2 \quad \Longrightarrow \quad x=\left(4-y^2\right)^2 \text {. }
$$
Note that the first equation implies $y \geq 0$, while the third one implies $4-y^2 \geq 0$. These restrictions should be observed before squaring the equations. The range is thus $[0,2]$.
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