이계도함수와 오목·볼록

a.

b. ${f}^{'}(x)=\frac{2ax{n}^{2}}{{({n}^{2}+{x}^{2})}^{2}}.$ When the amount of prey increases, the predator growth increases. c. ${f}^{''}(x)=\frac{2a{n}^{2}({n}^{2}-3{x}^{2})}{{({n}^{2}+{x}^{2})}^{3}}.$ When the amount of prey is extremely small, the rate at which predator growth is increasing is increasing, but when the amount of prey reaches above a certain threshold, the rate at which predator growth is increasing begins to decrease. d. At lower levels of prey, the prey is more easily able to avoid detection by the predator, so fewer prey individuals are consumed, resulting in less predator growth.

Since $f'(x)<0$, we need a decreasing function. This only applies to (ii), (iii), and (v). Since $f''(x)>0$, that means $f'(x)$ is increasing, so the slope of the function must be increasing. In (v), the slope is constant, so $f''(x)=0$–therefore, it's not (v). In (iii), the slope is decreasing, because near $a$ the curve is quite flat ($f'(x)$ near zero) but near $b$ the curve is very steeply decreasing ($f'(x)$ is a large negative number), so (iii) has a negative second derivative. By contrast, in (ii), the line starts out as steeply decreasing ($f'(x)$ is a strongly negative number) and becomes flatter and flatter ($f'(x)$ nears 0), so $f'(x)$ is increasing–in other words, $f''(x)>0$. So, (ii) is the only curve that has $f'(x)<0$ and $f''(x)>0$.

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.