Find where the function is increasing and where it is decreasing
f(x)=2sin(x) on [0,2pi]
Thanks
recall that f is increasing where f' > 0
f' = 2cosx
and you know that cosx is positive in QI and QIV. That is in (0,π/2) and (3π/2,2π).
To find where a function is increasing or decreasing, you need to analyze the sign of its derivative. Specifically, if the derivative is positive, the function is increasing, and if the derivative is negative, the function is decreasing.
To find the derivative of the function f(x) = 2sin(x), we use the chain rule.
First, let’s find the derivative of sin(x), which is cos(x).
Now, when finding the derivative of 2sin(x), we consider the 2 as a constant, so it doesn't affect the derivative. Therefore, the derivative of 2sin(x) is 2cos(x).
Now, let’s analyze the sign of the derivative in the interval [0, 2π].
Starting from x = 0, plug it into the derivative function to find f'(0):
f'(0) = 2cos(0) = 2(1) = 2
Since the derivative f'(0) is positive (greater than 0), this means that the function f(x) is increasing at x = 0.
Now, let’s move to x = π.
Plug it into the derivative function to find f'(π):
f'(π) = 2cos(π) = 2(-1) = -2
Here, the derivative f'(π) is negative (less than 0), indicating that the function f(x) is decreasing at x = π.
Lastly, look at x = 2π to determine if the function is increasing or decreasing.
Plug it into the derivative function to find f'(2π):
f'(2π) = 2cos(2π) = 2(1) = 2
At x = 2π, the derivative f'(2π) is positive (greater than 0), which means the function f(x) is increasing.
To summarize:
- The function f(x) = 2sin(x) is increasing on the interval [0, π].
- The function f(x) = 2sin(x) is decreasing on the interval [π, 2π].
- The function f(x) = 2sin(x) is increasing on the interval [2π, 2π] (but keep in mind that the interval [2π, 2π] is just a single point at x = 2π, so it doesn't have a clear trend of increasing or decreasing besides that particular point).