数学代写|MATH141 Calculus

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这是惠灵顿维多利亚大学微积分课程的代写成功案例。

MATH141课程简介

This course provides a thorough development of the differential calculus and an introduction to the integral calculus. The course builds on the ideas of functions and limits to define derivatives and integrals, as well as rules for computing these and applications to physical modelling.
This course is designed for in-person study, and it is strongly recommended that students attend lectures and tutorials on campus. In particular some assessment items will have a requirement of in-person attendance, although exceptions can be made under special circumstances.

Queries about any such exceptions can be sent to [email protected].

Prerequisites 

This course provides a thorough introduction to the differential calculus and an introduction to the integral calculus. Following a review of coordinate geometry and the use of equations to represent straight lines and circles, we introduce the concept of a function and many important examples including polynomial, rational, trigonometric, exponential and logarithm functions. The idea of a limit is central and leads to considering continuity and differentiability of a function. The definition of a derivative and rules for computing derviatives are deduced. Derivatives are applied to model physical problems. The course concludes by introducing the idea of an integral of a function and we demonstrate the fundamental property that integrals can be calculated using the inverse of differentiation.As well as developing methods of computation and applications of the calculus, the course focuses on its underlying concepts, the correct use of symbolic representation in mathematics and the importance of providing logical justification for its results and methods.

MATH141 Calculus HELP（EXAM HELP， ONLINE TUTOR）

Consider the function $f$ which is defined as the difference of the two sides, namely
$$
f(x)=x \sin x+x \cos x-1 .
$$
Being a composition of continuous functions, $f$ is then continuous and we also have
$$
f(0)=-1<0, \quad f(\pi / 2)=\frac{\pi}{2}-1=\frac{\pi-2}{2}>0 .
$$
In view of Bolzano’s theorem, this already implies that $f$ has a root $0<x<\pi / 2$.

Since $f$ is a polynomial on the intervals $(-\infty, 3)$ and $(3,+\infty)$, one easily finds that
$$
\begin{aligned}
& \lim {x \rightarrow 3^{-}} f(x)=\lim {x \rightarrow 3^{-}}\left(4 x^2+a x+b\right)=36+3 a+b, \
& \lim {x \rightarrow 3^{+}} f(x)=\lim {x \rightarrow 3^{+}}\left(2 x^3-b x+a\right)=54-3 b+a .
\end{aligned}
$$
In particular, the function $f$ is continuous at the given point if and only if
$$
36+3 a+b=54-3 b+a=a+b-2 .
$$
Solving this system of equations, we obtain a unique solution which is given by
$$
54-3 b=b-2 \quad \Longrightarrow \quad 4 b=56 \quad \Longrightarrow \quad b=14 \quad \Longrightarrow \quad a=-19 .
$$
In other words, $f$ is continuous at the given point if and only if $a=-19$ and $b=14$.

Being a polynomial, the given function is continuous and one can easily check that
$$
f(-2)=-29, \quad f(-1)=7, \quad f(0)=1, \quad f(1)=-5, \quad f(2)=31 .
$$
Since the values $f(-2)$ and $f(-1)$ have opposite signs, $f$ has a root that lies in $(-2,-1)$. The same argument yields a second root in $(0,1)$ and also a third root in $(1,2)$.

Since the first limit involves infinite values of $x$, it should be clear that
$$
L=\lim {x \rightarrow+\infty} \frac{2 x^4-4 x^2+5}{3 x^4-7 x+2}=\lim {x \rightarrow+\infty} \frac{2 x^4}{3 x^4}=\frac{2}{3} .
$$
For the second limit, the denominator becomes zero when $x=3$, while the numerator is nonzero at that point. Thus, one needs to factor the denominator and this gives
$$
M=\lim {x \rightarrow 3^{-}} \frac{x^3-5 x+4}{(x-3)\left(x^2+3 x+1\right)}=\lim {x \rightarrow 3^{-}} \frac{16}{19(x-3)}=-\infty
$$

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