이계도함수와 오목·볼록

We differentiate a few time to find the pattern.

$$\begin{aligned} \frac{\mathrm{d}}{\mathrm{d}x}\{2^x\}&=2^x\log 2\\ \frac{\mathrm{d}^2}{\mathrm{d}{x}^2}\{2^x\}&=2^x\log2 \cdot \log 2 = 2^x(\log2)^2\\ \frac{\mathrm{d}^3}{\mathrm{d}{x}^3}\{2^x\}&=2^x(\log2)^2 \cdot \log 2 = 2^x(\log2)^3 \\ \text{Every time we differentiate, we multiply the original function by another factor of $\log 2$. So, the $n$th derivative is given by:} \\ \frac{\mathrm{d}^n}{\mathrm{d}{x}^n}\{2^x\}&=2^x(\log2)^n \end{aligned}$$

We differentiate using the power rule.

$$\begin{aligned} \frac{\mathrm{d}f}{\mathrm{d}x}&=3ax^2+2bx+c\\ \frac{\mathrm{d}^2f}{\mathrm{d}{x}^2}&=6ax+2b\\ \frac{\mathrm{d}^3f}{\mathrm{d}{x}^3}&=6a\\ \frac{\mathrm{d}^4f}{\mathrm{d}{x}^4}&=0 \end{aligned}$$

In the above work, remember that $a$, $b$, $c$, and $d$ are all constants. Since they are nonzero constants, $\displaystyle \frac{\mathrm{d}^3f}{\mathrm{d}{x}^3} =6a\neq 0$. So, the fourth derivative is the first derivative to be identically zero: $n=4$.

Since $f(x)$ is differentiable over all real numbers, it is also continuous over all real numbers. Similarly, $f'(x)$ is differentiable over all real numbers, so it is also continuous over all real numbers, and so on for the first $n$ derivatives of $f(x)$.

Rolle's Theorem tells us that if $a$ and $b$ are distinct roots of a function $g$, then $g'(x)=0$ for some $c$ in $(a,b)$. That is, $g'$ has a root strictly between $a$ and $b$. Expanding this idea, if $g$ has $m+1$ distinct roots, then $g'$ must have at least $m$ distinct roots, as in the sketch below.

So, if $f^{(n)}(x)$ has only $m$ roots, then $f^{(n-1)}(x)$ has at most $m+1$ roots. Similarly, since $f^{(n-1)}(x)$ has at most $m+1$ roots, $f^{(n-2)}(x)$ has at most $m+2$ roots. Continuing in this way, we see $f(x)=f^{(n-n)}(x)$ has at most $m+n$ distinct roots.