Trig Functions — Question 2

Question 2 - Solution

(a) Analyze $g(x) = 2\tan\!\left(x - \frac{\pi}{4}\right) + 1$

• Period: $\boxed{\pi}$

• Phase shift: $\boxed{\frac{\pi}{4} \text{ to the right}}$

• Vertical shift: $\boxed{+1}$

Asymptotes occur when:

$$x - \frac{\pi}{4} = \frac{\pi}{2}+k\pi,\quad k\in\mathbb Z \Rightarrow x = \frac{3\pi}{4}+k\pi$$

So one period lies on:

$$\left(-\frac{\pi}{4},\ \frac{3\pi}{4}\right)$$

• All asymptotes: $\boxed{x=3\pi/4+k\pi,\ k\in\mathbb Z}$.

(b) Sketch one period

The intercepts in this period are $(0,-1)$ and $(\pi/4-\arctan(1/2),0)$.

Key features:

• Left asymptote: $x = -\frac{\pi}{4}$

• Right asymptote: $x = \frac{3\pi}{4}$

• Midpoint (inflection): $x = \frac{\pi}{4},\ g\!\left(\frac{\pi}{4}\right)=1$

(c) Solve $g(x) = 3$:

$$2\tan\!\left(x - \frac{\pi}{4}\right) + 1 = 3 \Rightarrow \tan\!\left(x - \frac{\pi}{4}\right) = 1$$

$$x - \frac{\pi}{4} = \frac{\pi}{4} + n\pi \Rightarrow x = \frac{\pi}{2} + n\pi$$

Values in $[0, 2\pi]$:

$$\boxed{x = \frac{\pi}{2},\ \frac{3\pi}{2}}$$