미분 전체 문제

(a) The function $f(x) = |x-2|$ is continuous at $a = 2$, because $f(2) = 0 = \lim_{x \to 2} f(x)$, but not differentiable at $a = 2$ because $f$ has a corner point at $a = 2$. (b) This is impossible. Any function that is differentiable at a point must have a limit at the point. (c) Let $f$ be the function defined to be $f(x) = 1$ for every value of $x \ne -2$, and such that $f(-2) = 4$. This function has a limit of $1$ as $x \to -2$, but since $f(-2) = 4 \ne 1$, the function is not continus at $a = -2$. (d) A possible function $p$ that has the desired properties is shown in the following figure.

(a) $g$ has neither a local maximum nor local minimum at $x = 2$ because $f'$ does not change sign at $x = 2$. (b) Because $g'(2) = 0$ and $g'$ is positive for $1 < x < 2$ and for $2 < x < 3$, it must be that $g'$ is decreasing for $1 < x < 2$ and $g'$ is increasing for $2 < x < 3$. Since $g'$ changes from decreasing to increasing at $x = 2$, it follows that $g'$ has a local minimum at $x = 2$ and thus $g''(2) = 0$. In addition, $g''(x)$ must change sign from negative to positive at $x = 2$. In particular, $g''$ is negative for $1 < x < 2$ and positive for $2 < x < 3$. (c) $g$ has a horizontal tangent line at $x = 2$, is increasing before $x = 2$ and decreasing after $x = 2$, and concave down before $x = 2$ and concave up after $x = 2$, and thus $g$ has a point of inflection at $x = 2$.

(a) Note that $p''(x)=0$ at $x=-1$ and $x=2$. The second derivative sign chart for $p$ shows that $p''(x)$ is negative for $-1 < x < 2$ and positive for all other values of $x$. Since $p''$ changes sign at $x = -1$ and $x = 2$, $p$ has points of inflection at $x = -1$ and $x = 2$. In particular, $p$ is concave down for $-1 < x < 2$ and concave up for all other values of $x$. (b) If we compute $p''\left(\frac{\sqrt{5}-1}{2} \right)$, we find that $p''\left(\frac{\sqrt{5}-1}{2} \right) \approx -1.205$, which means that $p$ is concave down at $x = \frac{\sqrt{5}-1}{2}$. Since $p$ has a horizontal tangent line at this point, it follows that $p$ has a local maximum at $x = \frac{\sqrt{5}-1}{2}$ by the second derivative test. (c) Since $p(2) = \frac{12}{e^2}$ and $p'(2) = -\frac{5}{e^2}$, the equation of the tangent line to $p$ at $x = 2$ is given by $y = -\frac{5}{e^2}(x-2) + \frac{12}{e^2}$. To determine whether the tangent line lies above or below the graph of $y=p(x)$ at $x = 2$, we consider the concavity of $p$. Recall we know that $p''(2) = 0$ and that $p$ is concave down to the left of $x = 2$ and concave up to the right. Thus, the tangent line lies neither solely below nor solely above the graph of $y = p(x)$ at $x = 2$; it lies below $y = p(x)$ to the right of $x = 2$ and above $y = p(x)$ to the left of $x = 2$.

(a) Writing $p(x) = x^3 - ax^2 = x^2(x-a)$, we see that $p(x) = 0$ at $x=0$ and $x=a$. Further, since $x^2 \ge 0$ for all $x$, we note that $p(x) \le 0$ whenever $x < a$ and $p(x) > 0$ for all $x > a$. A typical such function is plotted below. (b) To find the critical numbers of $p$, we first compute $p'(x)$. Treating $a$ as constant, we have $p'(x) = 3x^2 - 2ax = x(3x - 2a)$. Thus, setting $p'(x) = 0$, we find that $x(3x-2a) = 0$ and thus $x = 0$ and $x = \frac{2a}{3}$ are the critical numbers of $p$. (c) Observe that $p''(x) = 6x - 2a$ and thus $p''(x) = 0$ implies that $x = \frac{a}{3}$. Since $p''(x)$ is linear with positive slope, we see that $p''(x)$ changes sign from negative to positive at $x = \frac{a}{3}$, and thus $p$ is concave down for $x < \frac{a}{3}$ and concave up for $x > \frac{a}{3}$ and $x = \frac{a}{3}$ is the only inflection point of $p$. (d) There is always a critical number of $p$ at $x = 0$. But the other critical number at $x = \frac{2a}{3}$ and the inflection point at $x = \frac{a}{3}$ both obviously depend directly on $a$: as we increase the value of $a$, both the location of the critical number and the inflection point move to the right along with $a$.

(a) The value of the numerator at $x=c$ is $e^{-c} \ne 0$, while the value of the denominator is $0$ at $x=c$. Thus, $\lim_{x \to c^+} \frac{e^{-x}}{x-c} = \infty$ and $\lim_{x \to c^-} \frac{e^{-x}}{x-c} = -\infty$. This makes the line $x=c$ a vertical asymptote for $q$. (b) We can write $q(x)$ as $q(x) = \frac{1}{(x-c)e^x}$ and thus see the denominator increases without bound as $x \to \infty$. Therefore, $\lim_{x \to \infty} \frac{e^{-x}}{x-c} = 0$. This makes the line $y=0$ a horizontal asymptote for $q$. As $x \to -\infty$, $e^{-x}$ increases without bound, while $(x-c)$ decreases without bound. Thus, $q$ has an indeterminate form as $x \to -infty$. Applying L'Hopital's Rule gives us $$\lim_{x \to -\infty} \frac{e^{-x}}{x-c} = \lim_{x \to -\infty} \frac{-e^{-x}}{1} = -\infty\,.$$ (c) Using $q(x) = \frac{1}{(x-c)e^x}$, the quotient and product rules along with algebraic simplification show that $$q'(x) = -\frac{(x-c)e^x + e^x}{(x-c)^2\left(e^x\right)^2} = -\frac{x-c+1}{(x-c)^2e^x}\,.$$ The function $q$ has critical points in its domain where $q'(x) = 0$ or $q'(x)$ doesn't exist. Note that $q'(x)$ is undefined at $x=c$, but $c$ is not in the domain of mq. Thus, the only critical points of $q$ occur when $q'(x)=0$. The only way the fraction $q'(x)$ can be $0$ is if the numerator is $0$. Since $x-c+1=0$ when $x=c-1$, the only critical number for $q$ is $x=c-1$. (d) The values of the exponential function $e^x$ and the quadratic $(x-c)^2$ are positive for $x \neq c$, so the sign of $q'(x) = -\frac{x-c+1}{(x-c)^2e^x}$ depends entirely on the sign of $x-(c-1)$. When $x < c-1$ we see that $q'(x) > 0$ and when $x > c-1$ we have $q'(x) < 0$. Thus, $q$ is increasing on $(-\infty, c-1)$ and is decreasing on $(c-1,c)$, so it follows that $q(c-1)$ is a relative maximum value. This value is a local maximum, since the value of $q$ can be made as large as we like as shown in part (b). (e) We've shown that a typical member of the family of functions given by $q(x) = \frac{1}{(x-c)e^x}$ has a vertical asymptote at $x = c$ and a critical number at which a local maximum occurs one unit to the left of $c$ at $x = c-1$. A typical such graph is shown below. Note how $q$ is increasing and decreasing according to our earlier work in (d).

(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.