롤의 정리와 평균값 정리

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 $\left[-\left(\frac{\pi}{3}+\frac{\sqrt{3}}{2}\right),\left(\frac{\pi}{3}+\frac{\sqrt{3}}{2}\right)\right]$.

Define $h(x)=f(x)-g(x)$, and notice $h(a)=f(a)-g(a)<0$ and $h(b)=f(b)-g(b)>0$. Since $h$ is the difference of two functions that are continuous over $[a,b]$ and differentiable over $(a,b)$, also $h$ is continuous over $[a,b]$ and differentiable over $(a,b)$. So, by the Mean Value Theorem, there exists some $c \in (a,b)$ with $$h'(c)=\frac{h(b)-h(a)}{b-a}$$ Since $(a,b)$ is an interval, $b>a$, so the denominator of the above expression is positive; since $h(b)>0>h(a)$, also the numerator of the above expression is positive. So, $h'(c)>0$ for some $c \in (a,b)$. Since $h'(c)=f'(c)-g'(c)$, we conclude $f'(c)>g'(c)$ for some $c \in (a,b)$.

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.