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f(x) = sin^(4)x+cos^(4)x in [0,(pi)/(2)]...

`f(x) = sin^(4)x+cos^(4)x` in `[0,(pi)/(2)]`

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To solve the problem, we need to verify Rolle's theorem for the function \( f(x) = \sin^4 x + \cos^4 x \) on the interval \([0, \frac{\pi}{2}]\). ### Step 1: Check the conditions of Rolle's Theorem Rolle's theorem states that if a function is continuous on a closed interval \([a, b]\), differentiable on the open interval \((a, b)\), and \(f(a) = f(b)\), then there exists at least one \(c\) in \((a, b)\) such that \(f'(c) = 0\). 1. **Continuity**: The function \(f(x) = \sin^4 x + \cos^4 x\) is composed of sine and cosine functions, which are continuous everywhere. Therefore, \(f(x)\) is continuous on \([0, \frac{\pi}{2}]\). 2. **Differentiability**: The function \(f(x)\) is also differentiable on \((0, \frac{\pi}{2})\) since it is composed of differentiable functions. 3. **Equal endpoints**: We need to check if \(f(0) = f\left(\frac{\pi}{2}\right)\): - \(f(0) = \sin^4(0) + \cos^4(0) = 0 + 1 = 1\) - \(f\left(\frac{\pi}{2}\right) = \sin^4\left(\frac{\pi}{2}\right) + \cos^4\left(\frac{\pi}{2}\right) = 1 + 0 = 1\) Since \(f(0) = f\left(\frac{\pi}{2}\right) = 1\), all conditions of Rolle's theorem are satisfied. ### Step 2: Find the derivative \(f'(x)\) Next, we find the derivative of \(f(x)\): \[ f'(x) = \frac{d}{dx}(\sin^4 x + \cos^4 x) \] Using the chain rule: \[ f'(x) = 4\sin^3 x \cdot \cos x - 4\cos^3 x \cdot \sin x \] This can be factored as: \[ f'(x) = 4\sin x \cos x (\sin^2 x - \cos^2 x) \] ### Step 3: Set the derivative equal to zero To find \(c\), we set \(f'(x) = 0\): \[ 4\sin x \cos x (\sin^2 x - \cos^2 x) = 0 \] This gives us two cases: 1. \(\sin x = 0\) or \(\cos x = 0\) (which are not in the open interval \((0, \frac{\pi}{2})\)) 2. \(\sin^2 x - \cos^2 x = 0\) From \(\sin^2 x - \cos^2 x = 0\), we have: \[ \sin^2 x = \cos^2 x \implies \tan^2 x = 1 \implies \tan x = 1 \] Thus, \(x = \frac{\pi}{4}\). ### Step 4: Verify if \(c\) is in the interval Since \(\frac{\pi}{4} \in (0, \frac{\pi}{2})\), we have found a point \(c\) that satisfies the conditions of Rolle's theorem. ### Conclusion We have verified that all conditions of Rolle's theorem are satisfied for the function \(f(x) = \sin^4 x + \cos^4 x\) on the interval \([0, \frac{\pi}{2}]\), and we found \(c = \frac{\pi}{4}\). ---

To solve the problem, we need to verify Rolle's theorem for the function \( f(x) = \sin^4 x + \cos^4 x \) on the interval \([0, \frac{\pi}{2}]\). ### Step 1: Check the conditions of Rolle's Theorem Rolle's theorem states that if a function is continuous on a closed interval \([a, b]\), differentiable on the open interval \((a, b)\), and \(f(a) = f(b)\), then there exists at least one \(c\) in \((a, b)\) such that \(f'(c) = 0\). 1. **Continuity**: The function \(f(x) = \sin^4 x + \cos^4 x\) is composed of sine and cosine functions, which are continuous everywhere. Therefore, \(f(x)\) is continuous on \([0, \frac{\pi}{2}]\). 2. **Differentiability**: The function \(f(x)\) is also differentiable on \((0, \frac{\pi}{2})\) since it is composed of differentiable functions. ...
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