Final Answer:
The number of critical points for y = sin(x-π) is c) ₃.
Step-by-step explanation:
To find the critical points of the function y = sin(x-π), we need to determine where the derivative is equal to zero or undefined. The derivative of sin(x-π) with respect to x involves applying the chain rule. The derivative of sin(u) is cos(u) ⋅ du/dx, where u = x-π.
So, for y = sin(x-π), the derivative y' is cos(x-π). To find critical points, we set y' = cos(x-π) equal to zero and solve for x.
cos(x-π) = 0
Using the properties of cosine, we know that cos(π) = -1, so cos(x-π) is zero when x-π = π + kπ, where k is an integer. Solving for x, we get x = 2πk.
There are three critical points within a single period of the sine function, corresponding to k = -1, 0, 1. Therefore, the correct answer is c) ₃, indicating that there are three critical points for y = sin(x-π) within the specified interval.