Final answer:
The critical numbers of the function f(x) = 3sin(x)cos(x) on the interval (-π, π) are x = -π/4 and x = π/4. The function f(x) is increasing on the intervals (-π, -π/4) and (π/4, π) and decreasing on the interval (-π/4, π/4).
Step-by-step explanation:
The critical numbers of the function f(x) = 3sin(x)cos(x) on the interval (-π, π) can be found by finding the values of x where the derivative of f(x) equals zero or is undefined.
To find the derivative of f(x), we can use the product rule, which states that the derivative of the product of two functions is equal to the derivative of the first function times the second function, plus the first function times the derivative of the second function.
The derivative of f(x) = 3sin(x)cos(x) is f'(x) = 3cos^2(x) - 3sin^2(x).
Setting f'(x) = 0, we can solve for x to find the critical numbers. By factoring and using trigonometric identities, we get cos(2x) = 0, which gives us x = -π/4 and x = π/4 as critical numbers.
Therefore, the critical numbers of f are x = -π/4 and x = π/4.
To determine the intervals on which f is increasing or decreasing, we can use the first derivative test. We can evaluate the sign of the derivative f'(x) in the intervals determined by the critical numbers.
Since f'(x) = 3cos^2(x) - 3sin^2(x), we can analyze the sign of the expression in each interval to determine whether f is increasing or decreasing.
For x < -π/4, both sin^2(x) and cos^2(x) are positive, so f'(x) > 0. Therefore, f(x) is increasing in this interval.
For -π/4 < x < π/4, sin^2(x) is positive and cos^2(x) is negative. Since the expression 3cos^2(x) - 3sin^2(x) is negative in this interval, f(x) is decreasing in this interval.
For x > π/4, both sin^2(x) and cos^2(x) are positive again, so f'(x) > 0. Therefore, f(x) is increasing in this interval.
In summary, the critical numbers of f are x = -π/4 and x = π/4. The function f(x) is increasing in the interval (-π, -π/4) and (π/4, π), and decreasing in the interval (-π/4, π/4).