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HELPP! f(x) = cos x − sin x, [0, 2π] use MVT to find a point that satisfies c.
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use (f(b) - f(a)) / (b - a) = f'(c)
b would equal 2pi and a could equal 0
Wait do you mean for integrals or for derivatives?
what invn177 is doing, but the answer will be a (fraction *'s pi, a fraction * pi)
the practive problem was cos x + sinx so they had -cosx = -sinx but i dont know what sin x - cos x would equal
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wait how you get that answer..?
Let f(x) = cos x + sin x, a = 0, b = 2π. Then f'(x) = −sin x + cos x By the MVT, there exists c [0, 2π] such that−sin c + cos c = f'(c) = f(2π) − f(0)/ 2π − 0 = 0.Thus −sin c = −cos c, and c = 1/4π, 5/4π on interval [0, 2π]
thats the answer to the practice problem
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