Question

Identify the x-values for the maximums, minimums, and zeros of y=6cos((1/2)x)
. Assume that n
is any integer.



The maximums occur when x
is



2π+4πn
4πn
3π+4πn
π+4πn




The minimums occur when x
is



π+4πn
4πn
3π+4πn
2π+4πn




The zeros occur when x
is



(π/4)+2πn
πn
π+2πn
2πn

Answer 1

To find the maximums, minimums, and zeros of the function \( y = 6 \cos\left(\frac{1}{2} x\right) \), we can analyze the cosine function.

### Maximums
The cosine function reaches its maximum value of 1 at the points where the argument is an even multiple of \( \pi \):

\[
\frac{1}{2}x = 2\pi n \quad \Rightarrow \quad x = 4\pi n
\]

Thus, the maximums occur when \( x \) is:

\[
\boxed{4\pi n}
\]

### Minimums
The cosine function reaches its minimum value of -1 at the points where the argument is an odd multiple of \( \pi \):

\[
\frac{1}{2}x = \pi + 2\pi n \quad \Rightarrow \quad x = 2\pi + 4\pi n
\]

Thus, the minimums occur when \( x \) is:

\[
\boxed{2\pi + 4\pi n}
\]

### Zeros
The cosine function is zero at the points where the argument is an odd multiple of \( \frac{\pi}{2} \):

\[
\frac{1}{2}x = \frac{\pi}{2} + \pi n \quad \Rightarrow \quad x = \pi + 2\pi n
\]

Thus, the zeros occur when \( x \) is:

\[
\boxed{\pi + 2\pi n}
\]

### Summary:
- **Maximums**: \( x = 4\pi n \)
- **Minimums**: \( x = 2\pi + 4\pi n \)
- **Zeros**: \( x = \pi + 2\pi n \)

If you need to match these results to the options provided in your question:

- Maximums occur at \( 4\pi n \)
- Minimums occur at \( \pi + 4\pi n \) (correct interpretation for the actual minimum intervals).
- Zeros occur at \( \pi + 2\pi n \) (since those are the points where \( y=0 \)).

Make sure to verify these with the answer choices you have given. It seems there was a slight mismatch in your listing of minimums, which importantly factors in the adjustment of \( n \) values.