show the following linearization of the function f(x)=cos(3x)-3sin(3x) at x=2pi
To find the linearization of a function at a particular point, follow these steps:
1. Start with the given function: f(x) = cos(3x) - 3sin(3x).
2. Take the derivative of the function to find its slope: f'(x) = -3sin(3x) - 9cos(3x).
3. Evaluate the derivative at x = 2π: f'(2π) = -3sin(6π) - 9cos(6π).
Note: sin(6π) = 0 and cos(6π) = 1, simplifying the expression to f'(2π) = -9.
4. Use the slope from step 3 and the given point (x = 2π) to write the equation of the tangent line:
L(x) = f'(2π)(x - 2π) + f(2π).
Since f(2π) = cos(6π) - 3sin(6π), and both values simplify to 1, the equation becomes:
L(x) = -9(x - 2π) + 1.
Therefore, the linearization of the function f(x) = cos(3x) - 3sin(3x) at x = 2π is L(x) = -9(x - 2π) + 1.