Question:

Rolle Theorem f(x) = sin x + cos x. Find c ε [0,2,π]

Updated On: Apr 13, 2025
Hide Solution
collegedunia
Verified By Collegedunia

Solution and Explanation

Step 1: State Rolle's Theorem and the Function
To apply Rolle's theorem to the function \( f(x) = \sin x + \cos x \) in the interval \([0, 2\pi]\), we need to check the conditions of the theorem.

Step 2: Check Continuity
The function \( f(x) = \sin x + \cos x \) is continuous on the interval \([0, 2\pi]\) as both \( \sin x \) and \( \cos x \) are continuous functions.

Step 3: Check Differentiability
The function \( f(x) = \sin x + \cos x \) is differentiable on the interval \((0, 2\pi)\) as both \( \sin x \) and \( \cos x \) are differentiable functions.

Step 4: Check Function Values at Endpoints
We have: \[ f(0) = \sin 0 + \cos 0 = 0 + 1 = 1 \quad \text{and} \quad f(2\pi) = \sin (2\pi) + \cos (2\pi) = 0 + 1 = 1 \] Therefore, \( f(0) = f(2\pi) = 1 \).

Step 5: Verify Rolle's Theorem Conditions
Since all the conditions of Rolle's theorem are satisfied (continuity, differentiability, and \( f(0) = f(2\pi) \)), there exists at least one point \( c \) in the interval \( (0, 2\pi) \) such that \( f'(c) = 0 \).

Step 6: Find the Derivative of f(x)
To find the value of \( c \), we need to find the derivative of \( f(x) \) and set it equal to zero: \[ f'(x) = \cos x - \sin x \] This is the derivative of the function \( f(x) = \sin x + \cos x \).

Step 7: Set the Derivative to Zero and Solve for x
Setting \( f'(x) = 0 \), we have: \[ \cos x - \sin x = 0 \] Simplifying further: \[ \cos x = \sin x \] This equation is satisfied when \( x = \frac{\pi}{4} \) and \( x = \frac{5\pi}{4} \).

Step 8: Find the Value of c
Therefore, the values of \( c \) where the derivative is zero are: \[ c = \frac{\pi}{4} \quad \text{and} \quad c = \frac{5\pi}{4} \] The solution provided in the original text had an error in stating that \( c = \frac{3\pi}{4} \). The correct values of \( c \) are \( \frac{\pi}{4} \) and \( \frac{5\pi}{4} \).

Step 9: State the Conclusion
Therefore, in the interval \([0, 2\pi]\), there exist at least two points \( c = \frac{\pi}{4} \) and \( c = \frac{5\pi}{4} \) such that \( f'(c) = 0 \).
Was this answer helpful?
1
0

Concepts Used:

Continuity & Differentiability

Definition of Differentiability

f(x) is said to be differentiable at the point x = a, if the derivative f ‘(a) be at every point in its domain. It is given by

Differentiability

Definition of Continuity

Mathematically, a function is said to be continuous at a point x = a,  if

It is implicit that if the left-hand limit (L.H.L), right-hand limit (R.H.L), and the value of the function at x=a exist and these parameters are equal to each other, then the function f is said to be continuous at x=a.

Continuity

If the function is unspecified or does not exist, then we say that the function is discontinuous.