maximum and minimum of y = cos(x) + sin (x)on the interval [0,2pi]
thank you very much
You're welcome!
Let
f(x) = cos(x) + sin (x)
Calculate derivative f'(x) and equate to zero. Solve for x.
For each value x1 of the solution, verify
f''(x). f''(x)<0 = maximum, and f''(x)>0 = minimum.
There may be multiple maxima and minima, choose the extreme values.
would the max be sqrt(2) at x = pi/4, and the min be -sqrt(2) at x = 5pi/4?
Exactly! where tan(x)=1.
To find the maximum and minimum of the function y = cos(x) + sin(x) on the interval [0, 2π], we can start by finding the critical points and then evaluate the function at these points and at the endpoints of the interval.
1. Critical points:
To find the critical points, we need to find where the derivative of the function is equal to zero or undefined. Let's find the derivative of y = cos(x) + sin(x) first.
dy/dx = -sin(x) + cos(x)
Setting dy/dx = 0, we have:
-sin(x) + cos(x) = 0
Rearranging the equation, we get:
cos(x) = sin(x)
Dividing both sides by cos(x), we have:
1 = tan(x)
Taking the inverse tangent of both sides, we find:
x = π/4, 5π/4
2. Evaluate the function at critical points and endpoints:
Now let's plug in the critical points and endpoints into the original function, y = cos(x) + sin(x), and see which value gives us the maximum and minimum:
At x = 0:
y = cos(0) + sin(0) = 1 + 0 = 1
At x = π/4:
y = cos(π/4) + sin(π/4) = (√2/2) + (√2/2) = √2
At x = 5π/4:
y = cos(5π/4) + sin(5π/4) = (-√2/2) + (-√2/2) = -√2
At x = 2π:
y = cos(2π) + sin(2π) = 1 + 0 = 1
3. Conclusion:
The maximum value of y is √2, which occurs at x = π/4, and the minimum value of y is -√2, which occurs at x = 5π/4, within the interval [0, 2π].