롤의 정리와 평균값 정리

The function $f(x)$ is continuous and differentiable over all real numbers. If $a$ and $b$ are distinct roots of $f(x)$, then $f'(c)=0$ for some $c$ strictly between $a$ and $b$ (Rolle's Theorem). So, let's think about $f'(x)$. $$f'(x)=(4x+1)^3+1$$ This is simple enough that we can find its zero explicitly: $$(4x+1)^3 +1= 0 \quad \Leftrightarrow \quad (4x+1)^3 = -1 \quad \Leftrightarrow \quad 4x+1 = -1 \quad \Leftrightarrow \quad 4x=-2$$ Hence $f'(c)=0$ only when $c=\dfrac{-1}{2}$. That the derivative only has a single zero is very useful. It means (via Rolle's theorem) that if $f(x)$ has distinct roots $a$ and $b$ with $a<b$, then we must have $a < \dfrac{-1}{2} < b$. This also means that $f(x)$ cannot have 3 distinct roots $a,b$ and (say) $q$ with $a<b<q$, because then Rolle's theorem would imply that $f'(x)$ would have two zeros — one between $a$ and $b$ and another between $b$ and $q$.

We've learned that $f(x)$ has at most two roots, and we've learned something about where those roots can exist, if there are indeed two of them. But that means $f(x)$ could have 0, 1, or 2 roots.

It's not easy to find a root of $f(x)$ by inspection. But we can get a good enough picture of the graph of $y=f(x)$ to tell exactly how many roots there are, just by exploiting the following properties of $f(x)$ and $f'(x)$.

• As $x$ tends to $\pm \infty$, $f(x)$ tends to $+\infty$.
• The derivative $f'(x)=(4x+1)^3+1$ is negative for $x<-\frac{1}{2}$ and is positive for $x>-\frac{1}{2}$. That is, $f(x)$ is decreasing for $x<-\frac{1}{2}$ and increasing for $x>-\frac{1}{2}$.
• $f\left(-\frac{1}{2}\right)=\frac{1}{16}-\frac{1}{2}<0$.

This means the function must look something like the graph below:

except we are unsure of the locations of the $x$-intercepts. At present we only know that one is to the left of $x=-1/2$ and one is to the right.

Note what happens to $f(x)$ as $x$ increases from strongly negative values to strongly positive values.

• When $x$ is large and negative, $f(x)>0$.
• As $x$ increases, $f(x)$ decreases continuously until $x=-\frac{1}{2}$, where $f(x)<0$. In particular, since $f(-1)=\frac{65}{16}>0$ and $f\big(-\frac{1}{2}\big)<0$ and $f(x)$ is continuous, the intermediate value theorem guarantees that $f(x)$ takes the value zero for some $x$ between $-\frac{1}{2}$ and $-1$. More descriptively put, as $x$ increases from hugely negative numbers to $-\frac{1}{2}$, $f(x)$ passes through zero exactly once.
• As $x$ increases beyond $-\frac{1}{2}$, $f(x)$ increases continuously, starting negative and becoming very large and positive when $x$ becomes large and positive. In particular, since $f(0)=\frac{1}{16}>0$ and $f\big(-\frac{1}{2}\big)<0$ and $f(x)$ is continuous, the intermediate value theorem guarantees that $f(x)$ takes the value zero for some $x$ between $-\frac{1}{2}$ and $0$. So, as $x$ increases from $-\frac{1}{2}$ to near $+\infty$, $f(x)$ again passes through zero exactly once.

So $f(x)$ must have exactly two roots, one with $x<-\frac{1}{2}$ and one with $x>-\frac{1}{2}$.