미분 전체 문제

The domain of arctangent is all real numbers. The domain of the logarithm function is all positive numbers, and $1+x^2$ is positive for all $x$. So, the domain of $f(x)$ is all real numbers.

In order to find where $f(x)$ is increasing, we find where $f'(x)$ is positive. $$f'(x)= \frac 2 {1+x^2} - \frac{2x}{1+x^2} = \frac{2-2x}{1+x^2}$$ Since the denominator is always positive, $f(x)$ is increasing when when $2-2x>0$. We conclude that $f(x)$ is increasing on the open interval $(-\infty,1)$.

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 function is even if $f(-x)=f(x)$.

$$\begin{aligned} f(-x)&=\frac{(-x)^4-(-x)^6}{e^{(-x)^2}}\\ &=\frac{x^4-x^6}{e^{x^2}}\\ &=f(x) \end{aligned}$$

So, $f(x)$ is even.

For any real number $x$, we will show that $f(x)=f(x+4\pi)$.

$$\begin{aligned} f(x+4\pi)&=\sin(x+4\pi)+\cos\left(\frac{x+4\pi}{2}\right)\\ &=\sin(x+4\pi)+\cos\left(\frac{x}{2}+2\pi\right)\\ &=\sin(x)+\cos\left(\frac{x}{2}\right)\\ &=f(x) \end{aligned}$$

So, $f(x)$ is periodic.

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

(a) The change in population is $P(2000) - P(1980) = 197 800 - 181 843 = 15 957$ people. (b) Since there were $15 957$ people added to Grand Rapids over the 20 years from 1980 to 2000, in an average year the population grew by $\frac{15,957}{20} \approx 798$ people/year. (c) $\frac{f(b)-f(a)}{b-a}=\frac{\Delta f}{\Delta x}$ measures the slope of a secant line through the points $(a,f(a))$ and $(b,f(b))$. (d) $$AV_{[0,20]}=\frac{P(20)-P(0)}{20-0}=\frac{197 800 - 181 843}{20}=\frac{15 957}{20} \approx 798$$ people per year. (e) $$\begin{aligned}AV_{[5,10]} &=\frac{P(10) - P(5)}{10-5} = \frac{189 116.72 - 185 444.20}{10-5} \approx 734.50 \\ AV_{[5,9]} &= \frac{188 376.44 - 185 444.20}{9-5} \approx 733.06 \\ AV_{[5,8]} &= \frac{187 639.06 - 185,444.20}{8-5} \approx 731.62 \\ AV_{[5,7]} &= \frac{186 904.57 - 185,444.20}{7-5} \approx 730.19 \\ AV_{[5,6]} &= \frac{186 172.95 - 185,444.20}{6-5} \approx 728.7535\end{aligned}$$ The units on each of these quantities are “people per year”.

(a) First, $P(7)-P(0) = 115(1.014)^7-115(1.014)^0 \approx 0.1175$ billion people is the total population change. It follows that the average rate of change over this time period is $AV_{[0,7]}=\frac{0.1175}{7} \approx 0.01679$ billion people per year. The instantaneous rate of change of the population at $t = 7$ is $$P'(7) = \lim_{h \to 0} AV_{[7,7+h]} = \lim_{h \to 0} \frac{115(1.014)^{7+h}-115(1.014)^7}{h}\,,$$ and approximating this quantity with small values of $h$, it follows that $P'(7) \approx 0.1762$. Hence, the instantaneous rate of change at $t = 7$ is greater than the average rate of change of $P$on the interval $[0,7]$. (b) Using the formula for average rate of change, $$AV_{[19,29]}=\frac{1.15(1.014)^{29}-1.15(1.014)^{19}}{29-19} \approx 0.02234$$ billion people/year. (c) We will say that today's date is July 1, 2015, which means that $t = 22.5$, so we want to compute $P'(22.5)$. The limit definition tells us that $$P'(22.5) = \lim_{h \to 0} \frac{115(1.014)^{22.5+h}-115(1.014)^{22.5}}{h}\,.$$ Estimating this limit by using small values of $h$, we determine that $P'(22.5) \approx 0.02186$, which is measured in billions of people per year. (d) We want the equation of the tangent line that has slope $P'(22.5) \approx 0.02186$ and that passes through the point $(22.5, P(22.5))$. Using point-slope form, it follows that the line is given by $$y - 1.57236 = 0.02186(t-22.5)\,.$$