Question

Verify Rolle's theorem for the following function $f(x) = e^{-x} (\sin x-\cos x) \ \ \ x \epsilon [ \frac{\pi}{4},\frac{5\pi}{4} ]$

Answer 1

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Answer 2

Answer:

We have given      

$f(x) = e^{-x} (\sin x-\cos x)\ \ \ x \epsilon [ \frac{\pi}{4},\frac{5\pi}{4} ]$

Now differentiate f(x) w. r. t. x

$f'(x) = e^{-x} (cosx - (-sinx)) - e^{-x} (sinx - cosx)\\f'(x) = e^{-x} (cosx + sinx) - e^{-x}(sinx - cosx)\\f'(x) = e^{-x} cosx + e^{-x}sinx - e^{-x} sinx +e^{-x} cosx = 2e^{-x} cosx$

f(x) is differentiable on open interval(π/4, 5π/4) and continuous on closed interval[π/4, 5π/4]

Now,  

$f(a) = f(\pi/4) = e^{-\pi/4} (\sin \pi/4-\cos\pi/4) = e^{-\pi/4} (1/\sqrt 2 - 1\sqrt 2) = 0\\f(b) = f(\pi/4) = e^{-5 \pi/4} (\sin 5\pi/4-\cos5\pi/4) \\f(b)= e^{-5pi/4} (-1/\sqrt 2 -(- 1\sqrt 2)) = e^{-5pi/4} (-1/\sqrt 2 + 1\sqrt 2) = 0$

f(a) = f(b) = 0

Hence, function satisfy all the condition of the Rolle's Theorem.

Now we have to show that there exist some c∈(1, 4) such that f'(c) = 0

$f(x) = e^{-x} (\sin x-\cos x)$

$f'(x) =2e^{-x} cosx$

$f'(c) =2e^{-c} cosc$ = 0

cosc = 0

cosc = cos90° = cosπ/2

c = π/2

c = π/2 ∈[π/4, 5π/4]

Hence Rolle's theorem is verified.