f(x)= (1-sinx) interval (0≤x≤2π) find where the function is increasing decreasing..
i know that f`(x)=1-sinx
it is equal to zero when x=pi/2
im stuck there..
if f'(x) >0 (strictly increasing) between a and b, then f(x) is increasing in the interval [a,b].
if f'(x) <0 (strictly decreasing) between b and c, then f(x) is decreasing in the interval [b,c].
You will find that the point b is common to both intervals, this is normal, because increasing or decreasing at a point is defined with respect to another point on the given interval.
To determine where the function f(x) = 1 - sin(x) is increasing or decreasing, you can use the first derivative test.
1. Find the derivative of f(x). You correctly found that f'(x) = 1 - sin(x).
2. Set the derivative equal to zero and solve for x. You correctly found that f'(x) = 0 when x = pi/2.
3. Determine the intervals on the x-axis before and after the critical point (x = pi/2). You can do this by choosing test points within each interval and determining the sign of the derivative for those points.
To find the intervals, you can create a sign chart using the test points. Let's choose x = 0 and x = pi as the test points.
- For x = 0, substitute it into the derivative f'(x) = 1 - sin(x):
f'(0) = 1 - sin(0) = 1 - 0 = 1
So, the sign of the derivative at x = 0 is positive (+).
- For x = pi, substitute it into the derivative:
f'(pi) = 1 - sin(pi) = 1 - 0 = 1
So, the sign of the derivative at x = pi is positive (+).
Based on the test results, we can create a sign chart:
x < pi/2 | x = pi/2 | x > pi/2
-----------------------------------------------------
f'(x) (+) 0 (+)
Noting that x = pi/2 is the only critical point, we can observe the sign changes of the derivative from positive to zero at x = pi/2, and then from zero to positive for both intervals. This means that the function f(x) is increasing before reaching x = pi/2 and then continues to increase afterward.
Therefore, the function f(x) = 1 - sin(x) is increasing on the intervals (0, pi/2) and (pi/2, 2pi).
Note: The intervals are open because the inequality in the given interval (0 ≤ x ≤ 2π) uses strict inequalities.