To understand the transformations of the function \( f(x) = \cos(2x) \) in relation to its parent function \( \cos(x) \), we can analyze what happens when we modify the argument of the cosine function.
The parent function is \( \cos(x) \), which has a period of \( 2\pi \). When we have the function \( \cos(kx) \) where \( k \) is a positive constant, it affects the period of the cosine function. Specifically, the period of \( \cos(kx) \) is given by:
\[ \text{Period} = \frac{2\pi}{k} \]
In our case, we have \( k = 2 \). Therefore, the period of \( f(x) = \cos(2x) \) is:
\[ \text{Period} = \frac{2\pi}{2} = \pi \]
This means that the graph of \( f(x) = \cos(2x) \) completes one full cycle over an interval of \( \pi \), compared to the parent function \( \cos(x) \) which completes one full cycle over an interval of \( 2\pi \).
Since the period is halved, we can infer that the graph of \( \cos(2x) \) experiences a horizontal compression. Specifically, the graph is compressed by a factor of \( \frac{1}{2} \) because the frequency effectively doubles due to the \( 2x \) in the argument.
In conclusion, the transformation of \( \cos(x) \) to \( \cos(2x) \) is a horizontal compression by a factor of \( \frac{1}{2} \).
Therefore, the correct answer is:
C. The graph is a horizontal compression by a factor of \( \frac{1}{2} \).