함수의 그래프 그리기

(a) Treating $m$ and $s$ as constants, by the chain rule, $$E'(x) = e^{-\frac{(x-m)^2}{2s^2}} \cdot \frac{d}{dx}\left[ -\frac{(x-m)^2}{2s^2} \right] = e^{-\frac{(x-m)^2}{2s^2}} \cdot \left[ -\frac{2(x-m)}{2s^2} \right]$$ Note that $E'(x)$ is defined for every value of $x$, thus the only critical numbers occur where $E'(x) = 0$. Since $e$ raised to any power is never zero, the only way that $E'(x) = 0$ is for $\left[ -\frac{2(x-m)}{2s^2} \right] = 0$, which implies that $x = m$. Thus, $x = m$ is the only critical number of $E$. (b) As we noted in (a), $x = m$ is the only critical number of $E$. More over, since $e$ to any power is always positive, the sign of $E'(x)$ depends entirely on the expression $-\frac{2(x-m)}{2s^2} = -\frac{1}{2}(x-m)$. This linear function of $x$ has negative slope, and thus is positive for $x < m$ and negative for $x > m$, and thus $E'(x)$ is positive for $x < m$ and negative for $x > m$. Therefore, $E$ is increasing for $x < m$ and decreasing for $x > m$, with a local maximum at $x = m$. (c) From the given formula for $E''(x)$ and our knowledge that $e$ to any power is positive, the only way $E''(x) = 0$ is for $$\left(\frac{(x-m)^2 - s^2}{s^4} \right) = 0\,,$$ which implies that $(x-m)^2 = s^2$, and therefore $x-m = \pm s$, or $x = m \pm s$. These are the two values for which $E''(x) = 0$. (d) As $x \to \pm \infty$, we know that the quadratic function $-\frac{(x-m)^2}{2s^2}$ must decrease without bound because the quadratic function opens down. Thus, the power to which we are raising $e$ in the formula for $E(x)$ is decreasing without bound, which means that $E(x)$ must approach $0$ since raising $e$ to more and more negative numbers drives the result to $0$. Thus, $\lim_{x \to \infty} E(x) = \lim_{x \to -\infty} E(x) = 0$. (e) We've shown that $E$ has a single critical number at $x = m$ and two potential points of inflection at $x = m \pm s$, as well as that $E(x) \to 0$ as $x \to \pm \infty$. All of these properties are reflected in the following figure.