함수의 그래프 그리기

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

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

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 $x \to -\infty$.

Let's consider $g(x)=\tan(3x)$ and $h(x)=\sin(4x)$ separately. Recall that $\pi$ is the period of tangent.

$$\begin{aligned} \tan X &= \tan(X+\pi)&&\text{for every $X$ in the domain of $\tan X$} \\ \text{Replacing $X$ with $3x$:} \\ \tan(3x)&=\tan(3x+\pi)&&\text{for every $x$ in the domain of $\tan 3x$}\\ \tan(3x)&=\tan\left(3\left(x+\frac{\pi}{3}\right)\right)&&\text{for every $x$ in the domain of $\tan 3x$}\\ g(x)&=g\left(x+\frac{\pi}{3}\right)&&\text{for every $x$ in the domain of $\tan 3x$} \\ \text{So, the period of $g(x)=\tan(3x)$ is $\dfrac{\pi}{3}$.} \\ \\ \text{Similarly, $2\pi$ is the period of sine.} \\ \sin(X)&=\sin(X+2\pi)&&\text{for every $X$ in the domain of $\sin(X)$} \\ \text{Replacing $X$ with $4x$:} \\ \sin(4x)&=\sin(4x+2\pi)&&\text{for every $x$ in the domain of $\sin(4x)$}\\ \sin(4x)&=\sin\left(4\left(x+\frac{\pi}{2}\right)\right)&&\text{for every $x$ in the domain of $\sin(4x)$}\\ h(x)&=h\left(x+\frac{\pi}{2}\right)&&\text{for every $x$ in the domain of $\sin(4x)$} \end{aligned}$$

So, the period of $h(x)=\sin(4x)$ is $\dfrac{\pi}{2}$.

All together, $f(x)=g(x)+h(x)$ will repeat when both $g(x)$ and $h(x)$ repeat. The least common integer multiple of $\dfrac{\pi}{3}$ and $\dfrac{\pi}{2}$ is $\pi$. Since $g(x)$ repeats every $\dfrac{\pi}{3}$ units, and $h(x)$ repeats every $\dfrac{\pi}{2}$ units, they will not both repeat until we move $\pi$ units. So, the period of $f(x)$ is $\pi$.