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.)