还在面临微积分的学习挑战吗？别担心！我们的calculus-guide团队专业为您解决导数、积分、极限、微分方程等微积分的问题。我们拥有深厚的专业背景和丰富的经验，能帮您完成高水平的作业和论文，让您的学习之路一路顺风！
以下是一些我们可以帮助您解决的问题：
微积分基础：导数、积分、极限、连续性等概念。
微分学：微分方程、微分算子、微分几何等相关内容。
积分学：泰勒级数、黎曼积分、牛顿-莱布尼茨公式等知识点。
曲线和曲面：曲线的参数方程、曲面的参数化等内容。
级数：收敛性、发散性、级数运算等问题。
无论您面临的问题是什么，我们都会尽力为您提供专业的帮助，确保您的学习之旅顺利无阻！
(a) Find all $x$ such that $f(x) \leq 2$ where
$$
f(x)=-x^2+1 \quad f(x)=(x-1)^2 \quad f(x)=x^3
$$
Write your answers in interval notation and draw them on the graphs of the functions.
(b) Using the functions in part a, find all $x$ such that $|f(x)| \leq 2$. Write your answers in interval notation and draw them on the graphs of the functions.
(c) Can you find upper bounds for the functions in part a? That is, for each function $f$ is there a number $M$ such that for all $x, f(x) \leq M$ ?
(d) What about lower bounds for the functions in part a? That is, for each $f$ can you find a number $m$ such that for all $x, f(x) \geq m$ ?
(e) What about finding upper and lower bounds for these functions restricted to the interval $[-1,1]$ ? That is, for each $f$ can you find numbers $M$ and $m$ such that for all $x$ in $[-1,1], m \leq f(x) \leq M ?$
(f) True or false? If $M$ is an upper bound for the function $f$ and $M^{\prime}$ is an upper bound for the function $g$, then for all $x$ which are in the domains of both $f$ and g,
$$
|f(x)+g(x)| \leq M+M^{\prime} .
$$
(a) To find all $x$ such that $f(x) \leq 2$ for each function, we can set up the inequality and solve for $x$.
For the function $f(x)=-x^2+1$, we have $-x^2+1 \leq 2$, which simplifies to $x^2 \geq -1$. Since $x^2$ is always nonnegative, this inequality is satisfied for all $x$. Therefore, the solution is $(-\infty, \infty)$.
For the function $f(x)=(x-1)^2$, we have $(x-1)^2 \leq 2$, which simplifies to $(x-1)^2-2 \leq 0$. By solving this inequality, we find $-1 \leq x \leq 3$. Therefore, the solution is $[-1, 3]$.
For the function $f(x)=x^3$, we have $x^3 \leq 2$. Since the cube of any real number can be either positive or negative, there are no real solutions to this inequality. Therefore, the solution is the empty set, $\emptyset$.
The graphs of these functions can be drawn accordingly to visualize the solutions.
(b) To find all $x$ such that $|f(x)| \leq 2$ for each function, we consider the absolute value of each function.
For $f(x)=-x^2+1$, we have $|f(x)| = |-x^2+1|$. This absolute value expression is always nonnegative, so $|f(x)| \leq 2$ is satisfied for all $x$. The solution is $(-\infty, \infty)$.
For $f(x)=(x-1)^2$, we have $|f(x)| = |(x-1)^2| = (x-1)^2$. From part (a), we know that the solution to $(x-1)^2 \leq 2$ is $[-1, 3]$. Therefore, the solution to $|f(x)| \leq 2$ is $[-1, 3]$.
For $f(x)=x^3$, we have $|f(x)| = |x^3| = x^3$. Since the cube of any real number can be positive or negative, $|f(x)| \leq 2$ is not satisfied for any $x$. The solution is the empty set, $\emptyset$.
(c) Yes, we can find upper bounds for each function in part (a).
For $f(x)=-x^2+1$, we can see that the maximum value of $f(x)$ is $1$. Therefore, for all $x$, $f(x) \leq 1$.
For $f(x)=(x-1)^2$, the function is always nonnegative or zero. The maximum value is $0$, so for all $x$, $f(x) \leq 0$.
For $f(x)=x^3$, there is no upper bound. The function can take arbitrarily large positive or negative values as $x$ approaches infinity or negative infinity.
(d) For each function in part (a), we can find a lower bound.
For $f(x)=-x^2+1$, we can see that the minimum value of $f(x)$ is $-1$. Therefore, for all $x$, $f(x) \geq -1$.
For $f(x)=(x-1)^2$, the function is always nonnegative or zero. The minimum value is $
(b) Suppose $f(x)=x^2$ and $g(x)=\sin x$.
i. Write the functions in part a in terms of $f$ and $g$. (For example, if $h(x)=2 x^2$ you can write $h$ in terms of $f$ as $h(x)=2 f(x)$.) If you find more than one way of writing these functions in terms of $f$ and $g$, show that they are equivalent.
ii. How can you change the graph of $f$ to obtain the graphs of the first three functions? Use your work from part a to help you.
i. Writing the functions in part (a) in terms of $f$ and $g$:
The first function in part (a) is $f(x)=-x^2+1$. We can write this function in terms of $f$ and $g$ as $f(x) = -f(x) + 2f(x) + 1$. Here, $f(x) = -x^2$ and $g(x) = 1$.
The second function in part (a) is $f(x)=(x-1)^2$. We can write this function in terms of $f$ and $g$ as $f(x) = f(x) + g(x) – 1$. Here, $f(x) = x^2$ and $g(x) = -2x + 1$.
The third function in part (a) is $f(x)=x^3$. We can write this function in terms of $f$ and $g$ as $f(x) = f(x) – 3g(x)$. Here, $f(x) = x^3$ and $g(x) = -x^2$.
These alternative representations are equivalent to the original functions given in part (a).
ii. To change the graph of $f(x)=x^2$ to obtain the graphs of the first three functions, we can use the work from part (a):
For the first function $f(x)=-x^2+1$, we shift the graph of $f(x)$ downward by $1$ unit. This is equivalent to subtracting $1$ from $f(x)$.
For the second function $f(x)=(x-1)^2$, we shift the graph of $f(x)$ to the right by $1$ unit. This is equivalent to replacing $x$ with $(x-1)$ in $f(x)$.
For the third function $f(x)=x^3$, we can see that it already corresponds to the graph of $f(x)$.
By making these adjustments to the graph of $f(x)=x^2$, we obtain the graphs of the first three functions given in part (a).
E-mail: help-assignment@gmail.com 微信:shuxuejun
help-assignment™是一个服务全球中国留学生的专业代写公司
专注提供稳定可靠的北美、澳洲、英国代写服务
专注于数学，统计，金融，经济，计算机科学，物理的作业代写服务