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It is not a local maximum/minimum because ${f}^{'}$ does not change sign

False; for example, $y=\sqrt{x}.$

a. Concave up for $x>\frac{4}{3},$ concave down for $x<\frac{4}{3}$ b. Inflection point at $x=\frac{4}{3}$

True, examine derivative

Answers will vary.

Found everywhere.

T

T

T

F

Critical point: $x=0$ Points of inflection: $\pm b/\sqrt{3}$

Critical point: $x=-b/(2a)$ No points of inflection

Critical points: $x=\frac{n\pi/2-b}{a}$, where $n$ is an odd integer Points of inflection: $(n\pi-b)/a$, where $n$ is an integer.

Critical point: $x=(a+b)/2$ Points of inflection: none

$\frac{dy}{dx} = -x/y$, so the function is increasing in second and fourth quadrants, decreasing in the first and third quadrants.

$\frac{d^2y}{dx^2} = -1/y - x^2/y^3$, which is positive when $y<0$ and is negative when $y>0$. Hence the function is concave down in the first and second quadrants and concave up in the third and fourth quadrants.

In general, this is false. For example, the function $f(x) = \dfrac{x^2-9}{x^2-9}$ has no vertical asymptotes, because it is equal to 1 in every point in its domain (and is undefined when $x=\pm3$).

However, it is certainly possible that $f(x)$ has a vertical asymptote at $x=-3$. For example, $f(x)=\dfrac{1}{x^2-9}$ has a vertical asymptote at $x=-3$. More generally, if $g(x)$ is continuous and $g(-3)\ne 0$, then $f(x)$ has a vertical asympotote at $x=-3$.

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.)

Vertical asymptotes occur where the function blows up. In rational functions, this can only happen when the denominator goes to 0. In our case, the denominator is 0 when $x=3$, and in this case the numerator is $147$. That means that as $x$ gets closer and closer to 3, the numerator gets closer and closer to 147 while the denominator gets closer and closer to 0, so $|f(x)|$ grows without bound. That is, there is a vertical asymptote at $x=3$.

The horizontal asymptotes are found by taking the limits as $x$ goes to infinity and negative infinity. In our case, they are the same, so we condense our work.

$$\begin{aligned} \lim_{x \to \pm \infty}\dfrac{x(2x+1)(x-7)}{3x^3-81} &=\lim_{x \to \pm \infty}\dfrac{2x^3+ax^2+bx+c}{3x^3-81} \\ \text{where $a$, $b$, ad $c$ are some constants. Remember, for rational functions, you can figure out the end behaviour by looking only at the terms with the highest degree--the others won't matter, so we don't bother finding them. From here, we divide the numerator and denominator by the highest power of $x$ in the denominator, $x^3$.} \\ &=\lim_{x \to \pm \infty}\dfrac{2x^3+ax^2+bx+c}{3x^3-81}\left(\frac{\tfrac{1}{x^3}}{\tfrac{1}{x^3}}\right)\\ &=\lim_{x \to \pm \infty}\dfrac{2+\tfrac{a}{x}+\tfrac{b}{x^2}+\tfrac{c}{x^3}}{3-\tfrac{81}{x^3}}\\ &=\dfrac{2+0+0+0}{3-0}=\frac{2}{3} \end{aligned}$$

So there is a horizontal asymptote of $y=\dfrac{2}{3}$ both as $x \to \infty$ and as $x \to -\infty$.

Since $f(x)$ is continuous over all real numbers, it has no vertical asymptote.

To find the horizontal asymptotes, we evaluate $\displaystyle \lim_{x \to \pm \infty}f(x)$.

$$\begin{aligned} \lim_{x \to \infty}10^{3x-7}&=\underbrace{\lim_{X \to \infty}10^X}_{\text{let }X=3x-7}=\infty \\ \text{So, there's no horizontal asymptote as $x \to \infty$.} \\ \lim_{x \to -\infty}10^{3x-7}&=\underbrace{\lim_{X \to -\infty}10^{X}}_{\text{let }X=3x-7}\\ &=\underbrace{\lim_{X' \to \infty}10^{-X'}}_{\text{let }X'=-X}\\ &=\lim_{X' \to \infty}\frac{1}{10^{X'}}\\ &=0 \end{aligned}$$

That is, $y=0$ is a horizontal asymptote as $x \to -\infty$.