함수의 그래프 그리기

True, examine derivative

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 $\pi$.

(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 $|k|<\dfrac{4}{5\sqrt[4]{5}}$