The Tangent Function

For $a\tan(bx)$ the period is $\frac{\pi}{|b|}$. Here $b = \frac{1}{2}$, so the period is $\frac{\pi}{1/2} = 2\pi$. On the graph, the asymptotes at $x = -\pi$ and $x = \pi$ are $2\pi$ apart.

• Correct: $2\pi$: Right: $\frac{\pi}{1/2} = 2\pi$, the distance between the asymptotes at $-\pi$ and $\pi$.
• $\pi$: That is the period of $\tan x$; the factor $\frac{1}{2}$ inside stretches the graph to period $2\pi$.
• $4\pi$: That is $\frac{2\pi}{b}$, the sine and cosine rule; tangent repeats after $\frac{\pi}{b}$.
• $\frac{\pi}{2}$: This multiplies $\pi$ by $b$ instead of dividing by $b$.

As $x$ increases toward $\pi$, $\frac{x}{2}$ increases toward $\frac{\pi}{2}$, where $\sin$ is near $1$ and $\cos$ is a small positive number. So $\tan\frac{x}{2}$ grows without bound and $\lim_{x \to \pi^-} f(x) = \infty$; on the graph the branch rises along the dashed line.

• Correct: $\lim_{x \to \pi^-} f(x) = \infty$: Right: from the left, the branch climbs without bound beside the asymptote.
• $\lim_{x \to \pi^-} f(x) = -\infty$: That is the behavior from the right, $\lim_{x \to \pi^+} f(x) = -\infty$, where the next branch starts.
• $\lim_{x \to \pi^-} f(x) = 0$: $f$ is $0$ at $x = 0$ and $x = 2\pi$, not near the asymptote.
• $\lim_{x \to \pi^-} f(x) = 2$: $f(\frac{\pi}{2}) = 2$ is one point on the branch; near $\pi$ the outputs keep growing.

On $(0, \pi)$ the outputs rise from $0$ toward the asymptote, so $f$ is increasing. The graph bends upward (concave up): over equal steps the rises get bigger, for example $f$ goes up $2$ on $\left[0, \frac{\pi}{2}\right]$ and much more than $2$ on $\left[\frac{\pi}{2}, \pi\right)$. So the rate of change of $f$ is increasing.

• Correct: $f$ is increasing, and the rate of change of $f$ is increasing: Right: rising outputs, and each equal step rises more than the last (concave up).
• $f$ is increasing, and the rate of change of $f$ is decreasing: That describes a branch bending downward, like $f$ on $(-\pi, 0)$; on $(0, \pi)$ the graph bends upward.
• $f$ is decreasing, and the rate of change of $f$ is increasing: The outputs go up from $0$, so $f$ is increasing here, not decreasing.
• $f$ is increasing, and the rate of change of $f$ is constant: A constant rate of change means a straight line; this branch curves upward.

The average rate of change of $d$ on $\left[\frac{\pi}{6}, \frac{\pi}{3}\right]$ is the slope of the secant line: $\frac{d(\pi/3) - d(\pi/6)}{\pi/3 - \pi/6}$$= \frac{4\sqrt{3} - \frac{4\sqrt{3}}{3}}{\pi/6}$$= \frac{8\sqrt{3}}{3}\cdot\frac{6}{\pi}$$= \frac{16\sqrt{3}}{\pi}$ kilometers per radian, about $8.82$.

• Answer: $\frac{16\,\sqrt{3}}{\pi}$ ≈ 8.821 kilometers per radian

From the table, $d\left(\frac{\pi}{3}\right) = 4\sqrt{3}$ and $d\left(\frac{\pi}{6}\right) = \frac{4\sqrt{3}}{3}$, so the ratio is $\frac{4\sqrt{3}}{4\sqrt{3}/3} = 3$. Tangent is not linear: doubling the input does not double the output.

• Correct: No: $d\left(\frac{\pi}{3}\right) = 4\sqrt{3}$, which is $3$ times $d\left(\frac{\pi}{6}\right) = \frac{4\sqrt{3}}{3}$: Right: the ratio of the table values is $3$, not $2$.
• Yes: $d$ is proportional to $\theta$, so doubling $\theta$ doubles $d$: $d$ would be proportional to $\theta$ only if its graph were a line through the origin; the rates in the table are not constant.
• No: $d\left(\frac{\pi}{3}\right)$ is $4$ times $d\left(\frac{\pi}{6}\right)$, because $\tan$ squares the angle: The ratio from the table is $3$; tangent does not square its input.
• Yes: $\tan\left(2\cdot\frac{\pi}{6}\right) = 2\tan\left(\frac{\pi}{6}\right)$: $\tan\frac{\pi}{3} = \sqrt{3}$ and $2\tan\frac{\pi}{6} = \frac{2\sqrt{3}}{3}$, which differ.