Assignment-daixieTM为您提供杜克大学Duke University MATH 22 Introductory Calculus II微积分代写代考和辅导服务！
Instructions:
Calculus is a branch of mathematics that deals with the study of rates of change and the accumulation of small quantities. It is a fundamental tool in many areas of science, engineering, and economics.
There are two main branches of calculus: differential calculus and integral calculus. Differential calculus deals with the study of rates of change and slopes of curves, while integral calculus deals with the accumulation of small quantities and the calculation of areas and volumes.
Calculus is used in a wide range of fields, including physics, engineering, economics, biology, and computer science. Some of the applications of calculus include modeling the motion of objects, optimizing systems, predicting population growth, and designing control systems.
The study of calculus typically begins with a foundation in algebra, geometry, and trigonometry. The concepts and techniques of calculus are built upon this foundation, and students are introduced to topics such as limits, derivatives, integrals, and differential equations.
Let $f(x)=(x+4)^{-1 / 2}, g(x)=x^2+1, h(x)=(x-3)^{1 / 2}$, and $j(x)=x^{-1}$.
Then $(j \circ((g \circ h)-(g \circ f)))(5)=$______
First, we need to find $(g \circ h)(x)$ and $(g \circ f)(x)$:
$$(g \circ h)(x) = g(h(x)) = h(x)^2 + 1 = (x-3) + 1 = x-2$$
$$(g \circ f)(x) = g(f(x)) = (x+4)^{-1/2}+1$$
Now we can plug these expressions into the given function:
$$(j \circ((g \circ h)-(g \circ f)))(5) = j((g \circ h)(5) – (g \circ f)(5))$$
$$(g \circ h)(5) = 5-2 = 3$$
$$(g \circ f)(5) = (5+4)^{-1/2}+1 = \frac{1}{3} + 1 = \frac{4}{3}$$
So we have:
$$(j \circ((g \circ h)-(g \circ f)))(5) = j((g \circ h)(5) – (g \circ f)(5)) = j(3 – \frac{4}{3}) = j(\frac{5}{3}) = \frac{1}{\frac{5}{3}} = \frac{3}{5}$$
Therefore, $(j \circ((g \circ h)-(g \circ f)))(5) = \boxed{\frac{3}{5}}$.
Let $g(x)=3 x-2$. Find a function $f$ such that $(f \circ g)(x)=18 x^2-36 x+19$.
Answer: $f(x)=$______
We have $(f \circ g)(x) = f(g(x)) = 18x^2 – 36x + 19$. We want to find a function $f$ that satisfies this equation.
Let $y = g(x) = 3x – 2$. Then we can write $x$ in terms of $y$ as $x = \frac{y+2}{3}$.
Substituting this into the expression for $(f \circ g)(x)$, we get:
$$(f \circ g)(x) = f\left(g\left(\frac{y+2}{3}\right)\right) = f\left(\frac{3y+4}{3}\right) = 18\left(\frac{3y+4}{3}\right)^2 – 36\left(\frac{3y+4}{3}\right) + 19$$
Simplifying this expression, we get:
$$(f \circ g)(x) = 6y^2 – 24y + 19$$
So we can define $f(x) = 6x^2 – 24x + 19$.
Therefore, $(f \circ g)(x) = f(g(x)) = f(3x-2) = 18x^2 – 36x + 19$ as desired.
Let $f(x)=x^2+1$. Find a function $g$ such that $(f \circ g)(x)=2+\frac{2}{x}+\frac{1}{x^2}$.
Answer: $g(x)=$____
Let $g(x)=ax+b$. Then we have: $$(f \circ g)(x) = f(g(x)) = f(ax+b) = (ax+b)^2+1 = a^2x^2+2abx+b^2+1.$$ We want to find $a$ and $b$ such that $(f \circ g)(x)=2+\frac{2}{x}+\frac{1}{x^2}$. Equating coefficients, we have: \begin{align*} a^2 &= 1 \ 2ab &= \frac{2}{x} \ b^2+1 &= \frac{1}{x^2} \end{align*} From the first equation, we have $a=\pm1$. If $a=1$, then the second equation gives $b=\frac{1}{x^2}$, but then the third equation gives $b^2+1\geq1$, which contradicts with the given expression. Therefore, we must have $a=-1$. Then the second equation becomes $-2b=\frac{2}{x}$, or $b=-\frac{1}{x}$. Substituting this into the third equation, we get: $$\frac{1}{x^2}+1 = \frac{1}{x^2}+b^2 = \frac{1}{x^2}+\frac{1}{x^2}= \frac{2}{x^2}.$$ Therefore, $g(x)=-x+\frac{1}{x}$ and we can check that $(f \circ g)(x) = f(-x+\frac{1}{x}) = (-x+\frac{1}{x})^2+1 = 2+\frac{2}{x}+\frac{1}{x^2}$, as desired.
