롤의 정리와 평균값 정리

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.