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）

When it comes to $f(x)$, the possible rational roots are $\pm 1, \pm 2, \pm 1 / 3, \pm 2 / 3$. Checking these possibilities, one finds that $x=1, x=-2$ and $x=-1 / 3$ are all roots. According to the factor theorem, each of $x-1, x+2$ and $x+1 / 3$ is thus a factor and one has
$$
f(x)=3(x-1)(x+2)(x+1 / 3)=(x-1)(x+2)(3 x+1) .
$$
When it comes to $g(x)$, let us first clear denominators and write
$$
6 g(x)=6 x^3-7 x^2+1
$$
The only possible rational roots are $\pm 1, \pm 1 / 2, \pm 1 / 3, \pm 1 / 6$. Checking these possibilities, one finds that $x=1, x=1 / 2$ and $x=-1 / 3$ are all roots. It easily follows that
$$
6 g(x)=6(x-1)(x-1 / 2)(x+1 / 3) \quad \Longrightarrow \quad g(x)=(x-1)(x-1 / 2)(x+1 / 3) .
$$

Letting $x=\sin \theta$ for convenience, one finds that $4 x^2+4 x-3=0$ and
$$
x=\frac{-4 \pm \sqrt{16+4 \cdot 12}}{8}=\frac{-4 \pm 8}{8} \Longrightarrow x=\frac{1}{2},-\frac{3}{2} .
$$
Since $x=\sin \theta$ must lie between -1 and 1 , the only relevant solution is $x=\sin \theta=\frac{1}{2}$. In view of the graph of the sine function, there should be two angles $0 \leq \theta \leq 2 \pi$ that satisfy this condition. The first one is $\theta_1=\frac{\pi}{6}$ and the second one is $\theta_2=\pi-\frac{\pi}{6}=\frac{5 \pi}{6}$.

When it comes to the first case, one can easily check that
$$
3(y+1)=\log _2(2 x-6) \quad \Longleftrightarrow \quad 2^{3 y+3}=2 x-6 \quad \Longleftrightarrow \quad 2^{3 y+2}=x-3,
$$
so the inverse function is defined by $f^{-1}(y)=2^{3 y+2}+3$. When it comes to the second case,
$$
y=\frac{7 \cdot 5^x-3}{4 \cdot 5^x+2} \Longleftrightarrow 4 y \cdot 5^x+2 y=7 \cdot 5^x-3 \quad \Longleftrightarrow \quad 2 y+3=(7-4 y) \cdot 5^x
$$
and this gives $5^x=\frac{2 y+3}{7-4 y}$, so the inverse function is defined by $f^{-1}(y)=\log _5 \frac{2 y+3}{7-4 y}$.

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