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Question: Verify Rolle’s Theorem for the function \(f(x) = {e^{ - x}}\sin x,x \in [0,\pi ]\)....

Verify Rolle’s Theorem for the function f(x)=e−xsin⁡x,x∈[0,π]f(x) = {e^{ - x}}\sin x,x \in [0,\pi ].

Explanation

Solution

Verify each and every condition to be required to satisfy the Rolle’s theorem and prove those
conditions for this given function. Which means, prove that the given function is continuous and differentiable in a given intervalf(0)=f(π)f(0) = f(\pi ) and find the value of x such that f(x)=0 and prove that the value of x is in between 0 andπ\pi .

Complete step-by-step answer:
We know that Rolle's theorem states if a function f is continuous on the closed interval [a, b] and differentiable on the open interval (a, b) such that f(a) = f(b), then f′(x) = 0 for some x with a ≤ x ≤ b.

First we need to prove that the given function is continuous in the given interval

We know that if two functions are continuous, then the product if the two functions will also be a continuous function which implies as e−x and sin⁡x{e^{ - x}}\,and\,\sin xare continuous functions which says the given function i.e.
f(x)=e−xsin⁡x,x∈[0,π]f(x) = {e^{ - x}}\sin x,x \in [0,\pi ] is continuous

Which implies the given function is differentiable in (0,π)(0,\pi )
Now, we need to prove that f(0)=f(π)f(0) = f(\pi )
\eqalign{ & f(0) = {e^{ - 0}}\sin 0 = 1 \times 0 = 0 \cr & f(\pi ) = {e^{ - \pi }}\sin \pi = {e^{ - \pi }} \times 0 = 0 \cr}

⇒\Rightarrow f(0)=f(π)f(0) = f(\pi )

Now we need to find that value of c such that f′(c) = 0 and prove that c is in between a and b i.e. 0 and π\pi

f′(x)=ddx(e−xsin⁡x)=0f'(x) = \dfrac{d}{{dx}}({e^{ - x}}\sin x) = 0
We know that ddxf(a).f(b)=f(a)ddxf(b)+f(b)ddxf(a)\dfrac{d}{{dx}}\\{ f(a).f(b)\\} = f(a)\dfrac{d}{{dx}}f(b) + f(b)\dfrac{d}{{dx}}f(a)

\eqalign{ & \dfrac{d}{{dx}}({e^{ - x}}\sin x) = {e^{ - x}}\dfrac{d}{{dx}}(\sin x) + \sin x\dfrac{d}{{dx}}({e^{ - x}}) \cr & = {e^{ - x}}\cos x - {e^{ - x}}\sin x \cr & = {e^{ - x}}(\cos x - \sin x) \cr & \cr}
Now f′(c) = 0

⇒f′(c)=e−c(cos⁡c−sin⁡c)=0 \Rightarrow f'(c) = {e^{ - c}}(\cos c - \sin c) = 0

i.e. either cos c=sinc or e−c=0{e^{ - c}} = 0 ;

But e−c{e^{ - c}}will be 0 when c=infinity but it is not in our specified interval so the solution should be

\eqalign{ & \cos c = \sin c \cr & c = n\pi + \dfrac{\pi }{4}\,\,\,\forall n \in Z\,\, \cr}

Now substitute the value of n in the above equation as n=1,2,3… and we should also check the negative integers also.
Then we get the values of c as −∞........−3π4,π4,5π4...... - \infty ........ - \dfrac{{3\pi }}{4},\dfrac{\pi }{4},\dfrac{{5\pi }}{4}......
The only value of c which is in our given interval is \dfrac{\pi }{4}.

This implies the given function is satisfying all the conditions of Rolle’s theorem. So, Rolle’s theorem for the given function is verified.

Note: Make sure to revise the Rolle’s theorem and prove all the conditions without forgetting any and Do not make mistakes in the process of finding whether it is a continuous and differentiable function or not. Solve the derivative function clearly .