One root of the equation cos x - x + frac{1}{ | vedclass

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(A) Let $f(x) = \cos x - x + \frac{1}{2}$.Calculate the value of the function at the boundaries of the interval $\left( 0, \frac{\pi}{2} \right)$:$f(0

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$\left( 0, \frac{\pi}{2} \right)$

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(A) Let $f(x) = \cos x - x + \frac{1}{2}$.Calculate the value of the function at the boundaries of the interval $\left( 0, \frac{\pi}{2} \right)$:$f(0) = \cos(0) - 0 + \frac{1}{2} = 1 + 0.5 = 1.5 > 0$.$f\left( \frac{\pi}{2} \right) = \cos\left( \frac{\pi}{2} \right) - \frac{\pi}{2} + \frac{1}{2} = 0 - \frac{\pi}{2} + 0.5 = \frac{1 - \pi}{2} \approx \frac{1 - 3.14}{2} = -1.07 Since $f(0) > 0$ and $f\left( \frac{\pi}{2} \right) < 0$,by the Intermediate Value Theorem,there exists at least one root in the interval $\left( 0, \frac{\pi}{2} \right)$.

One root of the equation $\cos x - x + \frac{1}{2} = 0$ lies in the interval

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