Find the period for a sine function that has a maximum at (pi/14,1) and a minimum at (3pi/14,-1) And write the equation: f(x) = sin ()

Answer 1

To find the period of the sine function based on the given maximum and minimum points, let's first identify the properties of the sine function.

The standard sine function, $f(x)=A\sin (B(x-C))+D$, has:

• A vertical shift determined by $D$,
• Amplitude given by $A$,
• A horizontal shift determined by $C$,
• A period given by $\frac{2\pi }{B}$.

Step 1: Determine the period.

You provided a maximum at $(\frac{\pi }{14},1)$ and a minimum at $(\frac{3\pi }{14},-1)$.

• The distance between the maximum and minimum points is:

$$\text{Distance}=\frac{3\pi }{14}-\frac{\pi }{14}=\frac{2\pi }{14}=\frac{\pi }{7}$$

In a sine function, the distance from a maximum to the next minimum is a quarter of the period ($\frac{P}{4}$), thus, we can setup the following equation for the period $P$:

$$\frac{P}{4}=\frac{\pi }{7}$$

Step 2: Solve for the period $P$:

$$P=4\cdot \frac{\pi }{7}=\frac{4\pi }{7}$$

Step 3: Write the equation of the sine function.

The maximum value is 1, which suggests the amplitude $A=1$ (since sine ranges from -1 to 1), and the vertical shift $D=0$. Since we need the function to have a maximum at $x=\frac{\pi }{14}$, we need to adjust for the horizontal shift.

Considering a standard sine function has its first maximum at $x=\frac{\pi }{2}$, we would typically need to shift it to the left to have it peak earlier.

Using the formula for the function, we have:

$$f(x)=A\sin (B(x-C))+D$$

Substituting our values $A=1$, $D=0$, $P=\frac{4\pi }{7}$ gives us $B$:

$$B=\frac{2\pi }{P}=\frac{2\pi }{\frac{4\pi }{7}}=\frac{7}{2}$$

We need to find $C$. Since the maximum occurs at $x=\frac{\pi }{14}$, we can express $C$ as follows:

$$\text{Set }B(x-C)=\frac{\pi }{2}\text{ when }x=\frac{\pi }{14}$$

Thus,

$$\frac{7}{2}(\frac{\pi }{14}-C)=\frac{\pi }{2}$$

Solving for $C$:

$$\frac{\pi }{14}-C=\frac{\pi }{7}\cdot \frac{2}{7}=\frac{\pi }{7}\cdot \frac{1}{4}=\frac{\pi }{2}\cdot \frac{1}{4}=\frac{\pi }{28}$$ $$C=\frac{\pi }{14}-\frac{\pi }{28}=\frac{2\pi }{28}-\frac{\pi }{28}=\frac{\pi }{28}$$

So, plugging it all into our equation, we get:

$$f(x)=\sin \left(\frac{7}{2}(x-\frac{\pi }{28})\right)$$

In conclusion, the function can be written as:

$$f(x)=\sin (\frac{7}{2}x-\frac{7\pi }{56})$$

This final equation represents the sine function with the specified maximum and minimum values.