using the interval [0,2pi]
and f(x) = sinx + cosx, obtain c £ (0,2pi) that satisfies the conclusion of Rolle's theorem
where £ mean element of and C means number
show step
f(0) = f(2π) = 1
so the condition is satisfied.
So, now you want c such that f'(c)=0
f' = cosx - sinx
f' = 0 at x = π/4
π/4 is in the interval [0,2π], so ta-da!
To use Rolle's theorem, we need to show that the function satisfies three conditions:
1. Continuity: The function f(x) = sin(x) + cos(x) is continuous on the closed interval [0, 2π]. This is because sin(x) and cos(x) are both continuous functions on the entire real line, so their sum is also continuous.
2. Differentiability: The function f(x) = sin(x) + cos(x) is differentiable on the open interval (0, 2π). The derivative of f(x) is f'(x) = cos(x) - sin(x), which is also continuous on the interval (0, 2π) because both cos(x) and sin(x) are continuous functions.
3. Equal endpoints: We need to check if f(0) = f(2π). Evaluating f(x) at the endpoints, we have:
f(0) = sin(0) + cos(0) = 0 + 1 = 1,
f(2π) = sin(2π) + cos(2π) = 0 + 1 = 1.
Since f(0) = f(2π) = 1, the third condition is satisfied.
Now that we have verified all the conditions of Rolle's theorem, we can proceed to find the specific value c in the interval (0, 2π) that satisfies the conclusion of the theorem.
According to Rolle's theorem, if a function satisfies the three conditions, there exists at least one value c in the open interval (0, 2π) such that f'(c) = 0.
To find such a value c, we need to find where the derivative f'(x) equals zero:
f'(x) = cos(x) - sin(x) = 0.
To solve this equation, we set cos(x) equal to sin(x):
cos(x) = sin(x).
Dividing both sides by cos(x), we get:
1 = tan(x).
The solutions to this equation lie in the interval (0, 2π), where tan(x) equals 1.
The solution to tan(x) = 1 in the interval (0, 2π) is x = π/4.
Therefore, the value c that satisfies the conclusion of Rolle's theorem on the interval [0, 2π] for the function f(x) = sin(x) + cos(x) is c = π/4.