극한·미분 심화 전체 문제

(a) We can plug any number into the cosine function, and it will return a number in $[-1,1]$. The domain of $\arcsin x$ is $[-1,1]$, so any number we plug into cosine will give us a valid number to plug into arcsine. So, the domain of $f(x)$ is all real numbers.

(b) We can plug any number into the cosine function, and it will return a number in $[-1,1]$. The domain of $\mathop{\mathrm{arccsc}} x$ is $(-\infty,-1] \cup [1,\infty)$, so in order to have a valid number to plug into arccosecant, we need $\cos x = \pm 1$. That is, the domain of $g(x)$ is all values $x=n\pi$ for some integer $n$.

(c) The domain of arccosine is $[-1,1]$. The domain of sine is all real numbers, so no matter what number arccosine spits out, we can safely plug it into sine. So, the domain of $h(x)$ is $[-1,1]$.

The tangent line is horizontal when $0=y'=a-\sin x$. That is, when $a=\sin x$.

• If $|a|>1$, then there is no value of $x$ for which $a=\sin x$, so the curve has no horizontal tangent lines.
• If $|a| = 1$, then there are infinitely many solutions to $a=\sin x$, but only one solution in the interval $[-\pi,\pi]$: $x=\arcsin(a)=\arcsin(\pm1)=\pm\frac{\pi}{2}$. Then the values of $x$ for which $a=\sin x$ are $x=2\pi n +a \frac{\pi}{2}$ for any integer $n$.
• If $|a|<1$, then there are infinitely many solutions to $a=\sin x$. The solution in the interval $\left(-\frac{\pi}{2},\frac{\pi}{2}\right)$ is given by $x=\arcsin(a)$. The other solution in the interval $\left(-\pi,\pi\right)$ is given by $x=\pi-\arcsin(a)$, as shown in the unit circles below.

So, the values of $x$ for which $x=\sin a$ are $x=2\pi n+\arcsin(a)$ and $x=2\pi n + \pi - \arcsin (a)$ for any integer $n$.

Remark: when $a=1$, then $$2\pi n+\arcsin(a) = 2\pi n + \dfrac{\pi}{2}=2\pi n +\pi -\left(\dfrac {\pi}{2}\right)=2\pi n+\pi-\arcsin(a).$$ Similarly, when $a=-1$, $$2\pi n+\arcsin(a) = 2\pi n - \dfrac{\pi}{2}=2\pi (n-1) +\pi -\left(-\dfrac {\pi}{2}\right) =2\pi(n-1)+\pi-\arcsin(a).$$ So, if we try to use the descriptions in the third bullet point to describe points where the tangent line is horizontal when $|a|=1$, we get the correct points but each point is listed twice. This is why we separated the case $|a|=1$ from the case $|a|<1$.

The linear approximation is $L(x)=3x-9$. Since we're approximating at $x=5$, $f(5)=L(5)$, and $f'(5)=L'(5)$. However, there is no guarantee that $f(x)$ and $L(x)$ have the same value when $x \neq 5$. So:

(a) $f(5)=L(5)=6$

(b) $f'(5)=L'(5)=3$

(c) there is not enough information to find $f(0)$.

For any constant $a$, $f(a)=(2a+5)$, and $f'(a)=2$, so our approximation gives us $$f(x) \approx (2a+5)+2(x-a)=2x+5$$

Since $f(x)$ itself is a linear function, the linear approximation is actually just $f(x)$ itself. As a consequence, the linear approximation is perfectly accurate for all values of $x$.