Assignment-daixieTM为您提供杜克大学Duke University MATH 21 Introductory Calculus I微积分代写代考和辅导服务！
Instructions:
It seems like you are referring to the credit policy for Mathematics courses at Duke University, where credit may be awarded on the basis of national/international examinations in mathematics such as College Board, International Baccalaureate, and British Advanced Level. This credit is equivalent to Mathematics 111L as a prerequisite for certain courses, including Mathematics 112L.
Specifically, students entering Mathematics 112L in the fall must have taken either Mathematics 111L or Mathematics 106L at Duke, or they may have earned credit through one of the aforementioned examinations. This means that students who have taken and performed well on these examinations may be able to bypass the Mathematics 111L prerequisite and enroll directly in Mathematics 112L.
It’s important to note that the specific requirements and policies for earning credit through national/international examinations may vary by institution, so students should always check with their school’s registrar or academic advisors for the most up-to-date information.
Let g(x) = 3x − 2. Find a function f such that (f ◦ g)(x) = 18x 2 − 36x + 19
Answer: f(x) =______
We want to find a function f such that (f ◦ g)(x) = 18x^2 − 36x + 19, where g(x) = 3x − 2.
Let y = g(x), then we can write (f ◦ g)(x) as f(y) = 18y^2 − 36y + 19.
Substituting y = g(x), we get:
f(g(x)) = 18(3x−2)^2 − 36(3x−2) + 19
= 18(9x^2 − 12x + 4) − 108x + 73
= 162x^2 − 252x + 73
Therefore, the function f(x) = 18x^2 − 36x + 19 ◦ g(x) is:
f(x) = 162x^2 − 252x + 73.
Alternatively, we can also solve for f(y) directly:
Let y = g(x) = 3x − 2.
Then (f ◦ g)(x) = f(y) = 18y^2 − 36y + 19.
Substituting y = 3x − 2, we get:
f(x) = 18(3x − 2)^2 − 36(3x − 2) + 19
= 18(9x^2 − 12x + 4) − 108x + 73
= 162x^2 − 252x + 73.
Therefore, the function f(x) = 162x^2 − 252x + 73 is such that (f ◦ g)(x) = 18x^2 − 36x + 19.
Let $h(x)=\arctan x$ for $x \geq 0, g(x)=\cos x$, and $f(x)=\left(1-x^2\right)^{-1}$. Find a number $p$ such that $(f \circ g \circ h)(x)=1+x^p$. Answer: $p=$_____
We have $(f \circ g \circ h)(x) = f(g(h(x))) = f(\cos(\arctan(x))) = \frac{1}{1 – \cos^2(\arctan(x))}$.
To simplify this expression, we can use the identity $\cos^2(\theta) + \sin^2(\theta) = 1$ and substitute $\theta = \arctan(x)$ to get:
$$\cos^2(\arctan(x)) + \sin^2(\arctan(x)) = 1 \quad \implies \quad \cos^2(\arctan(x)) = \frac{1}{1+x^2}$$
Substituting this into the expression for $(f \circ g \circ h)(x)$, we get:
$$(f \circ g \circ h)(x) = \frac{1}{1 – \frac{1}{1+x^2}} = 1 + \frac{x^2}{1+x^2} = 1 + \frac{x^2+1-1}{x^2+1} = 1 + \frac{1}{1+\frac{1}{x^2}}$$
Therefore, $p=\boxed{2}$.
Let $f(x)=3 x^2+5 x+1$. Find a function $g$ such that $(f \circ g)(x)=3 x^4+6 x^3-4 x^2-7 x+3$.
Answer: $g(x)=$______
We want to find a function $g$ such that $(f \circ g)(x)=3 x^4+6 x^3-4 x^2-7 x+3$.
If we let $y=g(x)$, then this becomes $f(y) = 3x^4+6x^3-4x^2-7x+3$.
Substituting $f(x)=3x^2+5x+1$ into this equation, we get:
$$3y^2+5y+1=3x^4+6x^3-4x^2-7x+3$$
Equating the coefficients of like terms, we get:
\begin{align*} 3y^2 &= 3x^4 \ 5y &= 6x^3 – 7x \ 1 &= -4x^2 + 3 \end{align*}
From the third equation, we can solve for $x^2$ as $x^2=\frac{1}{4}(3-1) = \frac{1}{2}$, which gives us $x=\pm\sqrt{\frac{1}{2}}=\pm\frac{1}{\sqrt{2}}$.
Substituting $x=\frac{1}{\sqrt{2}}$ into the second equation, we get:
$$5y = 6\left(\frac{1}{\sqrt{2}}\right)^3 – 7\left(\frac{1}{\sqrt{2}}\right) = -\frac{1}{\sqrt{2}}$$
Solving for $y$, we get $y = -\frac{1}{\sqrt{2}}\cdot\frac{1}{5} = -\frac{1}{5\sqrt{2}}$.
Therefore, one possible function $g$ is $g(x)=-\frac{1}{5\sqrt{2}}$.
Substituting $x=-\frac{1}{\sqrt{2}}$ into the second equation, we get:
$$5y = 6\left(-\frac{1}{\sqrt{2}}\right)^3 + 7\left(-\frac{1}{\sqrt{2}}\right) = -\frac{1}{\sqrt{2}}$$
Solving for $y$, we get $y = -\frac{1}{\sqrt{2}}\cdot\frac{1}{5} = -\frac{1}{5\sqrt{2}}$.
Therefore, another possible function $g$ is $g(x)=-\frac{1}{5\sqrt{2}}$.
So, the function $g$ can take on different values at different points, and two possible functions $g$ are $g(x)=-\frac{1}{5\sqrt{2}}$ and $g(x)=\frac{-1}{5\sqrt{2}}$.
