Question

find the period and write the equation of the sine function that has a maximum at $\left( \frac{\pi}{14}, 1 \
ight)$ and a minimum at $\left( \frac{3\pi}{14}, -1 \
ight)$. (1 point)

period = $\square$

$f(x) = \sin(\square)$

Explanation:

Step1: Find the period

The distance between a maximum and the next minimum of a sine function is half of its period. The x - coordinates of the maximum and minimum are $\frac{\pi}{14}$ and $\frac{3\pi}{14}$ respectively. The distance between them is $\frac{3\pi}{14}-\frac{\pi}{14}=\frac{2\pi}{14}=\frac{\pi}{7}$. Since this is half of the period, the period $T$ is $2\times\frac{\pi}{7}=\frac{2\pi}{7}$.

Step2: Find the equation of the sine function

The general form of a sine function is $y = A\sin(Bx - C)+D$. The amplitude $A$ is the distance from the mid - line to the maximum (or minimum). Since the maximum is 1 and the minimum is - 1, the amplitude $A = 1$ (because $A=\frac{1 - (-1)}{2}=1$). The mid - line $D = 0$ (since the average of 1 and - 1 is 0).
The period $T=\frac{2\pi}{|B|}$, and we know $T = \frac{2\pi}{7}$, so $\frac{2\pi}{|B|}=\frac{2\pi}{7}$, which implies $|B| = 7$, so $B = 7$ (we can take the positive value for simplicity here).
Now, we need to find the phase shift. The maximum of $y=\sin(x)$ occurs at $x=\frac{\pi}{2}$. For our function $y=\sin(7x - C)$, we know that a maximum occurs at $x = \frac{\pi}{14}$. So we set $7\times\frac{\pi}{14}-C=\frac{\pi}{2}$.
Simplify $7\times\frac{\pi}{14}=\frac{\pi}{2}$. Then $\frac{\pi}{2}-C=\frac{\pi}{2}$, which implies $C = 0$.
So the equation of the sine function is $f(x)=\sin(7x)$.

Answer:

period = $\frac{2\pi}{7}$
$f(x)=\sin(7x)$