Problem 20
Question
Identify the amplitude and period of the function. Then graph the function and describe the graph of \(g\) as a transformation of the graph of its parent function. \(g(x)=\frac{1}{2} \cos 4 \pi x\)
Step-by-Step Solution
Verified Answer
The amplitude of the function is \(\frac{1}{2}\) and the period is \(\frac{1}{2}\). When graphing, the function starts at its peak values and oscillates between these values with a cycle that completes every \(\frac{1}{2}\) units in the \(x\)-axis. Compared to the parent function \(\cos(x)\), the function \(g(x) = \frac{1}{2} \cos 4\pi x\) has undergone a vertical compression by \(\frac{1}{2}\) and a horizontal compression by \(\frac{1}{2}\)
1Step 1: Identify the Amplitude
The amplitude of a function is the peak deviation of the function from its center position. For this function \(g(x)=\frac{1}{2} \cos 4 \pi x\), the amplitude is represented by the coefficient of the cosine function, \(A = \frac{1}{2}\). That is, the function oscillates between \(\frac{1}{2}\) and \(-\frac{1}{2}\).
2Step 2: Identify the Period
The period of a sinusoidal function is calculated as \(P = \frac{2\pi}{|B|}\), where \(B\) is the coefficient of \(x\). In this function, \(B = 4\pi\). Hence, substituting the value of \(B\) into the formula, we get \(P = \frac{2\pi}{4\pi} = \frac{1}{2}\). So, the period of \(g(x)\) is \(\frac{1}{2}\). Furthermore, it also means that the function completes one full cycle in \(\frac{1}{2}\) unit of the \(x\)-value.
3Step 3: Graph the Function
Now that we have the amplitude and the period, we can graph the function. Start at the origin \( (0,0) \), then draw a wave that reaches up to \( \frac{1}{2} \) and down to \( -\frac{1}{2} \), completing 1 cycle every \( \frac{1}{2} \) unit on the \( x \)-axis. Make sure the function behaves the same way as the general cosine function, starting at its peak value at the origin and then oscillating between its peak and trough values.
4Step 4: Describe the Transformation
The parent function of \(g(x)\) is \(\cos(x)\) which has an amplitude of 1 and a period of \(2\pi\). Comparing that to our function \(g(x) = \frac{1}{2} \cos 4\pi x\), it can be seen that the graph of \(g(x)\) is compressed vertically by a factor of \(\frac{1}{2}\) (Amplitude), and horizontally by a factor of \( \frac{1}{2}\) (Period). This is the transformation that has taken place.
Key Concepts
AmplitudePeriodTransformation of Functions
Amplitude
When we talk about amplitude in the context of trigonometric functions, we are essentially discussing how far the graph of the function stretches or compresses vertically. Amplitude is always a positive value and it represents the maximum height (and depth) of the waveform with respect to its central horizontal axis. In mathematical terms, for a cosine or sine function expressed as \(a \cos(bx)\) or \(a \sin(bx)\), the amplitude is the absolute value of \(a\), which means:
- If \(g(x) = \frac{1}{2} \cos 4\pi x\), then the amplitude is \(\frac{1}{2}\).
- This tells us that the wave of \(g(x)\) oscillates between \(\frac{1}{2}\) and \(-\frac{1}{2}\).
Period
In trigonometric functions, the period is the horizontal length of one complete cycle. It defines how quickly the function repeats itself. The formula for the period \(P\) of a function of the form \(a \cos(bx)\) or \(a \sin(bx)\) is given by:
- \(P = \frac{2\pi}{|b|}\)
- \(P = \frac{2\pi}{4\pi} = \frac{1}{2}\)
Transformation of Functions
A transformation in the mathematical sense occurs when a basic function is altered in terms of its amplitude, period, phase shift, or vertical shift. In the exercise at hand, the function \(g(x) = \frac{1}{2} \cos 4\pi x\) is a transformed version of the standard cosine function \(\cos(x)\). Let's see what kind of transformations occur here:
- Vertical Compression: The coefficient in front of the cosine function is \(\frac{1}{2}\). This means the graph is compressed vertically by a factor of \(\frac{1}{2}\).
- Horizontal Compression: The function's period is transformed by altering the \(b\) value, here as \(4\pi\), resulting in a quicker completion of cycles, specifically \(\frac{1}{2}\) times the original cycle length.
Other exercises in this chapter
Problem 19
Sketch the angle. Then find its reference angle. \(\frac{15 \pi}{4}\)
View solution Problem 19
\(g(x)=\sec 4 x\)
View solution Problem 20
ERROR ANALYSIS Describe and correct the error in fi nding csc ?, given that ? is an acute angle of a right triangle and cos ? = 7 —11 .
View solution Problem 20
Simplify the expression. \(\tan (x-2 \pi)\)
View solution