미분 전체 문제

We know that $y'=2x$. So, if we choose a point $(\alpha,\alpha^2)$ on the curve $y=x^2$, then the tangent line to the curve at that point has slope $2\alpha$. That is, the tangent line has equation

$$\begin{aligned} (y-\alpha^2)&=2\alpha(x-\alpha)\\ \text{simplified, } \qquad y&=(2\alpha)x-\alpha^2 \\ \text{So, if $(1,-3)$ is on the tangent line, then} \\ -3&=(2\alpha)(1)-\alpha^2\\ \iff\qquad 0&=\alpha^2-2\alpha-3\\ \iff\qquad 0&=(\alpha-3)(\alpha+1)\\ \iff\qquad \alpha&=3, \quad\text{or}\qquad \alpha=-1. \\ \text{So, the tangent lines $y=(2\alpha)x-\alpha^2$ are} \\ y&=6x-9 \quad\text{and}\quad y=-2x-1. \end{aligned}$$

We already know that $\frac{\mathrm{d}}{\mathrm{d}x}x^2=2x$. So the slope of $y=3x^2$ at $x=a$ is $6a$. The tangent line to $y=3x^2$ at $x=a, y=3a^2$ is $y-3a^2=6a(x-a)$. This tangent line passes through $(2,9)$ if

$$\begin{aligned} 9-3a^2&=6a(2-a)\\ 3a^2-12a+9&=0\\ a^2-4a+3&=0\\ (a-3)(a-1)&=0\\ \implies~~a&=1,3 \end{aligned}$$

The points are $(1,3),\ (3,27)$.

This limit represents the derivative computed at $x=100180$ of the function $f(x)=\sqrt{x}$. Since the derivative of $f(x)$ is $\dfrac{1}{2\sqrt{x}}$, then its value at $x=100180$ is exactly $\dfrac{1}{2\sqrt{100180}}$.

Using the product rule, $f'(x)=(2x)g(x)+x^2g'(x)$, so $f'(0)=0\cdot g(x)+0\cdot g'(x)=0$. (Since $g$ is differentiable, $g'$ exists.)

Denote by $m$ the slope of the common tangent, by $(x_1,y_1)$ the point of tangency with $y=x^2$, and by $(x_2,y_2)$ the point of tangency with $y=x^2-2x+2$. Then we must have $$y_1=x_1^2\qquad y_2=x_2^2-2x_2+2\qquad m=2x_1=2x_2-2=\frac{y_2-y_1}{x_2-x_1}$$ From the “$m$” equations we get $x_1=\frac{m}{2}$, $x_2=\frac{m}{2}+1$ and

$$\begin{aligned} m&=\frac{y_2-y_1}{x_2-x_1}\\ &=y_2-y_1\\ &=x_2^2-2x_2+2-x_1^2\\ \\& =(x_2-x_1)(x_2+x_1)-2(x_2-1)\\ &=\left(\frac{m}{2}+1-\frac{m}{2}\right)\left(\frac{m}{2}+1+\frac{m}{2}\right) -2\left(\frac{m}{2}+1-1\right) \\ &=(1)(m+1)-2\frac{m}{2}\\ &=1\\ \text{So,}~~~m&=1,\qquad x_1=\tfrac{1}{2},\qquad y_1=\frac{1}{4},\qquad x_2=\frac{3}{2},\qquad y_2=\frac{9}{4}-3+2=\frac{5}{4} \end{aligned}$$

An equation of the common tangent is $y=x-\frac{1}{4}$.

This limit represents the derivative computed at $x=2$ of the function $f(x)=x^{2015}$. To see this, simply use the definition of the derivative at $a=2$ with $f(x)=x^{2015}$:

$$\begin{aligned} \left.\frac{\mathrm{d}}{\mathrm{d}x}\{f(x)\}\right|_{a} &= \lim_{x \to a}\frac{f(x)-f(a)}{x-a}\\ \left.\frac{\mathrm{d}}{\mathrm{d}x}\{x^{2015}\}\right|_{2} &=\lim_{x\to2} \frac{x^{2015}-2^{2015}}{x-2} \end{aligned}$$

Since the derivative of $f(x)$ is $2015\cdot x^{2014}$, then its value at $x=2$ is exactly $2015\cdot 2^{2014}$.

If $2x+\sin x$ is one–to–one over an interval, it never takes the same value for two distinct numbers in that interval. By Rolle's Theorem, if $f(a)=f(b)$ for distinct $a$ and $b$, then $f'(c)=0$ for some $c$ between $a$ and $b$. However, $f'(x)=2+\cos x$, which is never zero. In fact, $f'(x)\ge 1$ for all $x$, so $f(x)$ is strictly increasing over its entire domain. Therefore, our function $f$ never takes the same value twice, so it is one–to–one over all the real numbers, $(-\infty,\infty)$.

When we define the inverse function $f^{-1}(x)$, the domain of $f$ is the range of $f^{-1}$, and vice-versa. In general, we might have to restrict the domain of $f$ (and hence the range of $f^{-1}$) to an interval where $f$ is one–to–one, but in our case, this isn't necessary. So, the range of $f^{-1}$ is $(-\infty,\infty)$ and the domain of $f^{-1}$ is the range of $f$: $(-\infty,\infty)$.

If $f(x)=\dfrac{x}{2}+\sin x$ is one–to–one over an interval, it never takes the same value for two distinct numbers in that interval. By Rolle's Theorem, if $f(a)=f(b)$ for distinct $a$ and $b$, then $f'(c)=0$ for some $c$ between $a$ and $b$. Since $f'(x)=\frac{1}{2}+\cos x$, $f'(x)=0$ when $x=2n\pi\pm\frac{2\pi}{3}$ for some integer $n$. So, in particular, if $a$ and $b$ are distinct numbers in the interval $\left[-\frac{2\pi}{3},\frac{2\pi}{3}\right]$, then for every $c$ strictly between $a$ and $b$, $f'(c) \neq 0$, so by Rolle's Theorem $f(a) \neq f(b)$. Therefore $f(x)$ is one–to–one on the interval $\left[-\frac{2\pi}{3},\frac{2\pi}{3}\right]$.

We should also show that the interval $\left[-\frac{2\pi}{3},\frac{2\pi}{3}\right]$ cannot be extended to a larger interval over which $f(x)$ is still one–to–one. Consider the derivative $f'(x) = \frac{1}{2}+\cos x$. For all $-\frac{2\pi}{3} < x < \frac{2\pi}{3}$, we have $\cos x > -\frac{1}{2}$ (sketch the graph of $\cos x$ yourself) so that $f'(x)>0$ and $f(x)$ is increasing. But at $x=\frac{2\pi}{3}$, $f'(x)=0$, and then for $x$ a bit bigger than $\frac{2\pi}{3}$ we have $\cos x < -\frac{1}{2}$ so that $f'(x)<0$ and $f(x)$ is decreasing. So the graph “reverses direction", and $f(x)$ repeats values. (See the graph of $y=f(x)$ below.) The same is true for $x$ a little smaller than $-\frac{2\pi}{3}.$

When we define the inverse function $f^{-1}(x)$, first we restrict $f$ to $\left[-\frac{2\pi}{3},\frac{2\pi}{3}\right]$. Then the range of $f^{-1}$ is also $\left[-\frac{2\pi}{3},\frac{2\pi}{3}\right]$. The domain of $f^{-1}$ is the range of $f$ over this interval, so $\left[-\left(\frac{\pi}{3}+\frac{\sqrt{3}}{2}\right),\left(\frac{\pi}{3}+\frac{\sqrt{3}}{2}\right)\right]$.

Since $f(x)$ is differentiable over all real numbers, it is also continuous over all real numbers. We claim that $f(x)$ cannot have four or more distinct roots. For every two distinct roots $a<b$, Rolle's Theorem tells us there is a $c \in (a,b)$ such that $f'(c)=0$: that is, $c$ is a root of $f'$. Since $f'$ has only two distinct roots, $f$ can have at most three distinct roots.

We are asked to find the number of solutions to the equation $x^2+5x+1 = -\sin x$. The figure below contains the graphs of $y=x^2+5x+1$ and $y=-\sin x$. The solutions to $x^2+5x+1 = -\sin x$ are precisely the $x$'s where $y=x^2+5x+1$ and $y=-\sin x$ cross.

From the figure, it sure looks like there are two crossings. Since the function $-\sin x$ has range $[-1,1]$, if the two functions cross, then also $-1 \leq x^2+5x+1\leq 1$. This portion of the quadratic function is highlighted in blue in the figure. The $x$ coordinates of the end points of the blue arcs are found by solving $x^2+5x+1 = \pm1$, i.e. $x=0$, $-5$, and (using the quadratic equation) $x= \frac{-5\pm\sqrt{17}}{2}$.

We are now in a position to exploit the intuition that we have built using the above figure to write a concise argument showing that $f(x)$ has exactly two roots. Remember that, in general, if we want to show that a function has $n$ roots, we need to show that there exist $n$ distinct roots somewhere, and that there do not exist $n+1$ distinct roots. This argument is given below, in blue text.

• $f(x)$ is continuous over all real numbers
• $f(x)=0$ only when $-\sin x=x^2+5x+1$, which only happens when $|x^2+5x+1| \leq 1$. Thus, $f(x)$ only has roots in the intervals $\left[-5,\frac{-5-\sqrt{17}}{2}\right]$ and $\left[\frac{-5+\sqrt{17}}{2},0\right]$.
• $f(-5)=\sin(-5)+1 > 0$, and $f\left(\frac{-5-\sqrt{17}}{2}\right)=\sin\left(\frac{-5-\sqrt{17}}{2}\right)-1 <0$. So, by the IVT, $f(c)=0$ for some $c \in \left(-5,\frac{-5-\sqrt{17}}{2}\right)$.
• $f(0)=1>0$, and $f\left(\frac{-5+\sqrt{17}}{2}\right)=\sin\left(\frac{-5+\sqrt{17}}{2}\right)-1<0$. So, by the IVT, $f(c)=0$ for some $c \in \left(\frac{-5+\sqrt{17}}{2},0\right)$.
• $f'(x)=\cos x + 2x + 5$. If $f'(x)=0$, then $2x+5=-\cos x$, so $|2x+5| \leq 1$. So, the only interval that can contain roots of $f'(x)$ is $[-3,-2]$.
• Suppose $f(x)$ has more than two roots. Then it has two roots in the interval $\left[-5,\frac{-5-\sqrt{17}}{2}\right]$ OR it has two roots in the interval $\left[\frac{-5+\sqrt{17}}{2},0\right]$. Since $f(x)$ is differentiable for all real numbers, Rolle's Theorem tells us that $f'(x)$ has a root in $\left(-5,\frac{-5-\sqrt{17}}{2}\right)$ or in $\left(\frac{-5+\sqrt{17}}{2},0\right)$. However, since all roots of $f'(x)$ are in the interval $[-3,-2]$, and this interval shares no points with $\left(-5,\frac{-5-\sqrt{17}}{2}\right)$ or $\left(\frac{-5+\sqrt{17}}{2},0\right)$, this cannot be the case. Therefore $f(x)$ does not have more than two roots.

• Since $f(x)$ has at least two roots, and not more than two roots, $f(x)$ has exactly two roots.