Question

Verify Rolle's theorem for the function f (x) = e^x(sinx -cosx) in [π/4, 5π/4]​

Answer 1

Rolle's theorem

Statement :- Let f:[a,b] -> R be a function having the following three properties :-

1. Function is continuous in [a,b].

2. Function is differentiable in (a,b).

3. f(a) = f(b).

Then , there exist at least one c in (a,b) such that f'(c) = 0.

• Rolle's theorem is just sufficient to get Rolle's result. It is not a necessary condition. It implies that a function may give a point c where f'(c) = 0 even though the three conditions of Rolle's theorem doesn't satisfy. Basically, Rolle's theorem is like, if the three conditions hold then we will definitely get Rolle's result. If not, we can't say anything about the function.

• Any of the above conditions can not be relaxed.

• Note that differentially at end point is not necessary to get the Rolle's result.

Solution

1) Since, it is a nice function. It is continuous in [π/4,5π].

2) It is also differentiable in (π/4,5π/4).

3) Also, f(π/4) = 0 and f(5π/4) = 0 . Means f(π/4) = f(5π/4).

Thus, the conditions of Rolle's theorem are satisfied. There exists at least one c in (π/4,5π/4) such that f'(c) = 0.

f'(x) = e^x(sinx-cosx) + e^x(cosx+sinx)

f'(x) = 2e^xsinx which is zero at x = π between (π/4,5π/4)

Hence, verified

Answer 2

Basic Concept and Theory :-

Statement of Rolle's Theorem

This theorem states that

• Let f(x) be a real valued function defined on [a,b] such that

• 1. f(x) is continuous on [a, b].

• 2. f(x) is differentiable on (a, b).

• 3. f(a) = f(b).

This implies, there exist atleast one real number c belongs to (a, b) such that f'(c) = 0.

Remark :-

• Every sinx, cosx and Exponential functions are always continuous as well as differentiable on its domain.

Formulas of differentiation

$\red{ \rm : \implies \:\dfrac{d}{dx} sinx \: = \: cosx}$

$\red{ \rm : \implies \:\dfrac{d}{dx} cosx \: = - \: sinx}$

$\red{ \rm : \implies \:\dfrac{d}{dx} {e}^{x} \: = \: {e}^{x} }$

$\red{ \rm : \implies \:\dfrac{d}{dx} (uv) \: = u \: \dfrac{d}{dx} v \: + \: v\dfrac{d}{dx} u}$

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$\large\underline\purple{\bold{Solution :- }}$

Given

$\rm : \implies \:f(x) = {e}^{x} (sinx - cosx) \: on \: [\dfrac{\pi}{4} , \dfrac{5\pi}{4} ].$

$\bigstar \: \: \boxed{ \rm \: \pink{f(x) \: is \: continuous \: on \: [\dfrac{\pi}{4} , \dfrac{5\pi}{4} ]}}$

Now,

Differentiate f(x) w. r. t. x, we get

$\rm \: f'(x) \: = {e}^{x} \dfrac{d}{dx} (sinx - cosx) + (sinx - cosx)\dfrac{d}{dx} {e}^{x}$

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

$\rm : \implies \:f'(x) \: = \: {2e}^{x} sinx$

$\bigstar \: \: \boxed{ \rm \: \green{f(x) \: is \: differentiable \: on \: (\dfrac{\pi}{4} , \dfrac{5\pi}{4} )}}$

Now,

Consider,

$\rm : \implies \:f(\dfrac{\pi}{4} ) = {e}^{ \frac{\pi}{4} } (sin\dfrac{\pi}{4} - cos\dfrac{\pi}{4} )$

$\rm : \implies \:f(\dfrac{\pi}{4} ) = {e}^{ \frac{\pi}{4} } (\dfrac{1}{ \sqrt{2} } - \dfrac{1}{ \sqrt{2} } )$

$\bigstar \: \: \boxed{ \rm \: \pink{ \rm : \implies \:f(\dfrac{\pi}{4} ) = 0}}$

Consider,

$\rm : \implies \:f(\dfrac{5\pi}{4} ) = {e}^{ \frac{5\pi}{4} } (sin\dfrac{5\pi}{4} - cos\dfrac{5\pi}{4} )$

$\rm : \implies \:f(\dfrac{5\pi}{4} ) = {e}^{ \frac{5\pi}{4} } (sin(\pi \: + \dfrac{\pi}{4}) - cos(\pi + \dfrac{\pi}{4} ))$

$\rm : \implies \:f(\dfrac{5\pi}{4} ) = {e}^{ \frac{5\pi}{4} } ( - sin\dfrac{\pi}{4} + cos\dfrac{\pi}{4} )$

$\rm : \implies \:f(\dfrac{5\pi}{4} ) = {e}^{ \frac{5\pi}{4} } ( - \dfrac{1}{ \sqrt{2} } + \dfrac{1}{ \sqrt{2} } )$

$\bigstar \: \: \boxed{ \rm \: \pink{ \rm : \implies \:f(\dfrac{5\pi}{4} ) = 0}}$

This implies,

$\bigstar \: \: \boxed{ \rm \: \pink{ \rm : \implies \:f(\dfrac{\pi}{4} ) = f(\dfrac{5\pi}{4} )}}$

Hence,

All the three conditions of Rolle's Theorem are satisfied.

Therefore,

$\rm : there \: exist \: atleast \: one \: real \: number \: c \: \in \: (\dfrac{\pi}{4} , \dfrac{5\pi}{4}) \:$

such that f'(c) = 0

$\rm : \implies \:f'(c) \: = {2e}^{c} sinc = 0$

$\rm : \implies \:sinc \: = \: 0 \: \: as \: {e}^{c} \: \ne \: 0$

$\bigstar \: \: \boxed{ \pink{ \rm : \implies \:c \: = \: \pi \: \in \: (\dfrac{\pi}{4} , \dfrac{5\pi}{4} )}}$

$\bigstar \: \large\boxed{ \purple{ \bf \: Hence, \: Rolle's \: Theorem \: is \: Verified \: }}$