함수의 그래프 그리기

Since $x^2+1$ and $x^2+4$ are always positive, $f(x)$ and $h(x)$ are defined over all real numbers. So, $f(x)$ and $h(x)$ correspond to $A(x)$ and $B(x)$. Which is which? $A(0)=1=f(0)$ while $B(0)=2=h(0)$, so $A(x)=f(x)$ and $B(x)=h(x)$.

That leaves $g(x)$ and $k(x)$ matching to $C(x)$ and $D(x)$. The domain of $g(x)$ is all $x$ such that $x^2-1\ge0$. That is, $|x|\ge1$, like $C(x)$. The domain of $k(x)$ is all $x$ such that $x^2-4\ge0$. That is, $|x|\ge2$, like $D(x)$. So, $C(x)=g(x)$ and $D(x)=k(x)$.

(a) Since $f(0)=2$, we solve

$$\begin{aligned} 2&=\sqrt{\log^2(0+p)}\\ &=\sqrt{\log^2 p}\\ &=\left|\log p\right|\\ \log p &= \pm 2\\ p&=e^{\pm 2}\\ p&=e^2 \text{ or } p=\frac{1}{e^2} \end{aligned}$$

We know that $p$ is $e^2$ or $\dfrac{1}{e^2}$, but we have to decide between the two. In both cases, $f(0)=2$. Let's consider the domain of $f(x)$. Since $\log^2(x+p)$ is never negative, the square root does not restrict our domain. However, we can only take the logarithm of positive numbers. Therefore, the domain is

$$\begin{aligned} x \text{ such that } &x+p>0\\ x \text{ such that } &x>-p\\ \end{aligned}$$

If $p=\dfrac{1}{e^2}$, then the domain of $f(x)$ is $\left(-\dfrac{1}{e^2},\infty\right)$. In particular, since $-\dfrac{1}{e^2}>-1$, the domain of $f(x)$ does not include $x=-1$. However, it is clear from the graph that $f(-1)$ exists. So, $p=e^2$.

(b) Now, we need to figure out what $b$ is. Notice that $b$ is the end of the domain of $f(x)$, which we already found to be $(-p,\infty)$. So, $b=-p=-e^2$.

(As a quick check, if we take $e\approx 2.7$, then $-e^2=-7.29$, and this looks about right on the graph.)

(c) The $x$-intercept is the value of $x$ for which $f(x)=0$:

$$\begin{aligned} 0&=\sqrt{\log^2(x+p)}\\ 0&=\log(x+p)\\ 1&=x+p\\ x&=1-p=1-e^2 \end{aligned}$$

The $x$-intercept is $1-e^2$.

(As another quick check, the $x$-intercept we found is a distance of 1 from the vertical asymptote, and this looks about right on the graph.)

This function is symmetric across the $y$-axis, so it is even.

The function is not even, because it is not mirrored across the $y$-axis.

Assuming it continues as shown, the function is periodic, because the unit shown below is repeated:

Additionally, $f(x)$ is odd. In a function with odd symmetry, if we mirror the right-hand portion of the curve (the portion to the right of the $y$-axis) across both the $y$-axis and the $x$-axis, it lines up with the left-hand portion of the curve.

Since reflecting the right-hand portion of the graph across the $y$-axis, then the $x$-axis, gives us $f(x)$, we conclude $f(x)$ is odd.

Since the function is even, we simply reflect the portion shown across the $y$-axis to complete the sketch.

Since the function is odd, to complete the sketch, we reflect the portion shown across the $y$-axis (shown dashed), then the $x$-axis (shown in red).

(a) First, we differentiate. $$f(x) = x^5-x\qquad f'(x)=5x^4-1\qquad f''(x)=20x^3\qquad$$ The function and its first derivative tells us the following:

• $\displaystyle \lim_{x \to \infty} f(x)=\infty$, $\displaystyle \lim_{x \to -\infty} f(x)=-\infty$
• $f'(x)> 0$ (i.e. $f$ is increasing) for $|x|> \dfrac{1}{\sqrt[4]{5}}$
• $f'(x)=0$ (i.e. $f$ has critical points) for $x= \pm\dfrac{1}{\sqrt[4]{5}} \approx \pm 0.67$
• $f'(x)< 0$ (i.e. $f$ is decreasing) for $|x| < \dfrac{1}{\sqrt[4]{5}}$
• $f\left(\pm\dfrac{1}{\sqrt[4]{5}}\right)=\mp\dfrac{4}{5\sqrt[4]{5}}\approx\mp0.53$

This gives us a first idea of the shape of the graph.

We refine this skeleton using information from the second derivative.

• $f''(x)> 0$ (i.e. $f$ is concave up) for $x> 0$,
• $f''(x)=0$ (i.e. $f$ has an inflection point) for $x=0$, and
• $f''(x)< 0$ (i.e. $f$ is concave down) for $x<0$

Thus

• $f$ has no asymptotes
• $f$ has a local maximum at $x=-\dfrac{1}{\sqrt[4]{5}}$ and a local minimum at $x=\dfrac{1}{\sqrt[4]{5}}$
• $f$ has an inflection point at $x=0$
• $f$ is concave down for $x<0$ and concave up for $x>0$

(b) The function $x^5-x+k$ has a root at $x=x_0$ if and only if $x^5-x=-k$ at $x=x_0$. So the number of distinct real roots of $x^5-x+k$ is the number of times the curve $y=x^5-x$ crosses the horizontal line $y=-k$. The local maximum of $x^5-x$ (when $x=-\dfrac{1}{\sqrt[4]{5}}$) is $\dfrac{4}{5\sqrt[4]{5}}$, and the local minimum of $x^5-x$ (when $x=\dfrac{1}{\sqrt[4]{5}}$) is $-\dfrac{4}{5\sqrt[4]{5}}$. So, looking at the graph of $x^5-x$ above, we see that the number of distinct real roots of $x^5-x+k$ is

• 1 when $|k|>\dfrac{4}{5\sqrt[4]{5}}$
• 2 when $|k|=\dfrac{4}{5\sqrt[4]{5}}$
• 3 when $|k|<\dfrac{4}{5\sqrt[4]{5}}$

(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).