Find the intervals on which sinx-root3(cosx) is increasing and decreasing
Very urgent
does root3(cosx) mean cube root? If so, then
y = sinx - (cosx)^(1/3)
y is increasing when y' > 0
y' = cosx - 1/3 (cosx)^(-2/3) * (-sinx)
The graph is at
https://www.desmos.com/calculator/nwct6fs2np
I'm surprised at the abrupt changes of direction, since cosx is smooth at odd multiples of pi/2.
The graph of y' is at
https://www.wolframalpha.com/input/?i=cosx+-+1%2F3+%28cosx%29%5E%28-2%2F3%29+*+%28-sinx%29
you can see that y'=0 once every 2pi. y is decreasing/increasing in between the zeroes of y'
No its not cube root
Sinx - (root3)(cosx)
To find the intervals on which the function sin(x) - √3*cos(x) is increasing and decreasing, we need to analyze the derivative of the function. The derivative will help us determine where the function is increasing or decreasing.
First, let's find the derivative of the given function:
f(x) = sin(x) - √3*cos(x)
Taking the derivative with respect to x:
f'(x) = cos(x) + √3*sin(x)
Now, we can analyze the derivative to identify where the function is increasing or decreasing. In other words, we need to find the critical points and test intervals between them.
Critical points occur where the derivative equals zero or is undefined. So, we need to solve the equation f'(x) = 0:
cos(x) + √3*sin(x) = 0
Divide both sides by cos(x):
1 + √3*tan(x) = 0
√3*tan(x) = -1
tan(x) = -1/√3
To find the solutions for x, we take the inverse tangent (tan^(-1)) of both sides:
x = tan^(-1)(-1/√3)
This results in two solutions for x: -π/6 and -5π/6.
Now that we have the critical points, we can test the intervals between them and outside them to determine if the function is increasing or decreasing.
We know that sin(x) - √3*cos(x) is continuously differentiable (i.e., the derivative exists) for all real numbers, so we only need to test the intervals.
1. For x < -5π/6: Choose any value less than -5π/6, e.g., x = -2π. Substitute into f'(x) = cos(x) + √3*sin(x).
f'(-2π) ≈ cos(-2π) + √3*sin(-2π) ≈ 1 + 0 ≈ 1 > 0
Since f'(-2π) > 0, the function is increasing on the interval x < -5π/6.
2. For -5π/6 < x < -π/6: Choose any value between -5π/6 and -π/6, e.g., x = -π/2. Substitute into f'(x) = cos(x) + √3*sin(x).
f'(-π/2) ≈ cos(-π/2) + √3*sin(-π/2) ≈ 0 + (-√3) ≈ -√3 < 0
Since f'(-π/2) < 0, the function is decreasing on the interval -5π/6 < x < -π/6.
3. For -π/6 < x < π/6: Choose any value between -π/6 and π/6, e.g., x = 0. Substitute into f'(x) = cos(x) + √3*sin(x).
f'(0) ≈ cos(0) + √3*sin(0) ≈ 1 + 0 ≈ 1 > 0
Since f'(0) > 0, the function is increasing on the interval -π/6 < x < π/6.
4. For π/6 < x < 5π/6: Choose any value between π/6 and 5π/6, e.g., x = π/2. Substitute into f'(x) = cos(x) + √3*sin(x).
f'(π/2) ≈ cos(π/2) + √3*sin(π/2) ≈ 0 + √3 ≈ √3 > 0
Since f'(π/2) > 0, the function is increasing on the interval π/6 < x < 5π/6.
5. For x > 5π/6: Choose any value greater than 5π/6, e.g., x = 2π. Substitute into f'(x) = cos(x) + √3*sin(x).
f'(2π) ≈ cos(2π) + √3*sin(2π) ≈ 1 + 0 ≈ 1 > 0
Since f'(2π) > 0, the function is increasing on the interval x > 5π/6.
Summary:
The function sin(x) - √3*cos(x) is increasing on the intervals x < -5π/6, -π/6 < x < π/6, and x > 5π/6. The function is decreasing on the interval -5π/6 < x < -π/6 and π/6 < x < 5π/6.