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

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Question

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

Vedclass · Accepted Answer

$[0, \frac{\pi}{2}]$

Vedclass · Answer

(A) Let $f(x) = \cos x - x + \frac{1}{2}$.Calculate the value of the function at the boundaries of the interval $[0, \frac{\pi}{2}]$:$f(0) = \cos(0) - 0 + \frac{1}{2} = 1 + 0.5 = 1.5 > 0$.$f(\frac{\pi}{2}) = \cos(\frac{\pi}{2}) - \frac{\pi}{2} + \frac{1}{2} = 0 - \frac{\pi}{2} + 0.5 = \frac{1 - \pi}{2}$.Since $\pi \approx 3.14$,$1 - 3.14 = -2.14$,so $f(\frac{\pi}{2}) By the Intermediate Value Theorem,since $f(0) > 0$ and $f(\frac{\pi}{2}) < 0$,there must exist at least one root in the interval $[0, \frac{\pi}{2}]$.

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

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