# Verify Rolle’s Theorem for the function $f(x)={{x}^{2}}+2x-8,x\in \left[ -4,2 \right]$.

Last updated date: 27th Mar 2023

•

Total views: 307.2k

•

Views today: 5.85k

Answer

Verified

307.2k+ views

Hint: Use the basic definition of Rolle’s theorem that if any function f(x) is continuous in [a,b] and differentiable in (a,b) and further f(a) = f(b) then there exist ‘c’ such as ${{f}^{'}}\left( c \right)=0$ where $c\in \left( a,b \right)$

Complete step-by-step answer:

Rolle’s Theorem can be defined with any function f(x) where $x\in \left[ a,b \right]$ as

(i) If f(x) is continuous in [a,b] and

(ii) If f(x) is differentiable in (a,b) and

(iii) If f(a) = f(b) then ${{f}^{'}}\left( c \right)=0$ where $c\in \left[ a,b \right]$

Now, here we have

$f\left( x \right)={{x}^{2}}+2x-8$ and $x\in \left[ -4,2 \right]$

As a given function f(x) is a polynomial function, so we don’t need to worry about the continuity and differentiability of the function because any polynomial function will always be continuous and differentiable for $x\in R$.

Hence, the given function $f\left( x \right)={{x}^{2}}+2x-8$ is continuous in [a,b] and differentiable in (a,b).

Now, coming to the third step of Rolle’s theorem as given in the starting of the solution, we need to calculate f(-4) and f(2) to verify f(-4) = f(2).

So, we have

$\begin{align}

& f\left( x \right)={{x}^{2}}+2x-8 \\

& f\left( -4 \right)={{\left( -4 \right)}^{2}}+2\left( -4 \right)-8 \\

& f\left( -4 \right)=16-8-8=16-16 \\

\end{align}$

Or

f(-4) = 0 …………………(i)

And f(2) can be calculated as

$\begin{align}

& f\left( 2 \right)={{\left( 2 \right)}^{2}}+2\left( 2 \right)-8 \\

& f(2)=4+4-8=8-8 \\

\end{align}$

Or

f(2) = 0 ……………..(ii)

Hence, from equation (i) and (ii) we get,

f(-4) = f(2) = 0

So, it is verified all three points for Rolle’s Theorem.

Now, for ${{f}^{'}}\left( c \right)$ , we have

$f\left( x \right)={{x}^{2}}+2x-8$

Differentiating f(x) w.r.t ‘x’, we get

${{f}^{'}}\left( x \right)=2x+1...........\left( iii \right)$

Where we know $\dfrac{d}{dx}\left( {{x}^{n}} \right)=n{{x}^{n-1}}$

Now, for equation (iii), we have

${{f}^{'}}\left( c \right)=2c+1$

Equating above equation to ‘0’, we can get value of ‘c’ as

$\begin{align}

& {{f}^{'}}\left( c \right)=0 \\

& \Rightarrow 2c+1=0 \\

& c=\dfrac{-1}{2} \\

\end{align}$

Where $c\in \left[ -4,2 \right]$

Hence, the given function $f\left( x \right)={{x}^{2}}+2x-8$ with $x\in \left( -4,2 \right)$ is satisfying Rolle’s theorem as

f(x) is continuous and differentiable in [-4,2] and (-4,2) respectively and f(-4) = f(2) = 0, then

${{f}^{'}}\left( \dfrac{-1}{2} \right)=0$ where $\dfrac{-1}{2}$ is lying between [-4,2].

Note: One can get confused between Rolle’s Theorem and mean value theorem. So, be careful with these kinds of questions and be clear with both the theorems.

One can factorize the given relation as, (x+4) (x-2) where x = -4 and x = 2 are roots of the given equation. Hence, we can directly write f(-4) = f(2) = 0 without putting values of x to a given function.

Complete step-by-step answer:

Rolle’s Theorem can be defined with any function f(x) where $x\in \left[ a,b \right]$ as

(i) If f(x) is continuous in [a,b] and

(ii) If f(x) is differentiable in (a,b) and

(iii) If f(a) = f(b) then ${{f}^{'}}\left( c \right)=0$ where $c\in \left[ a,b \right]$

Now, here we have

$f\left( x \right)={{x}^{2}}+2x-8$ and $x\in \left[ -4,2 \right]$

As a given function f(x) is a polynomial function, so we don’t need to worry about the continuity and differentiability of the function because any polynomial function will always be continuous and differentiable for $x\in R$.

Hence, the given function $f\left( x \right)={{x}^{2}}+2x-8$ is continuous in [a,b] and differentiable in (a,b).

Now, coming to the third step of Rolle’s theorem as given in the starting of the solution, we need to calculate f(-4) and f(2) to verify f(-4) = f(2).

So, we have

$\begin{align}

& f\left( x \right)={{x}^{2}}+2x-8 \\

& f\left( -4 \right)={{\left( -4 \right)}^{2}}+2\left( -4 \right)-8 \\

& f\left( -4 \right)=16-8-8=16-16 \\

\end{align}$

Or

f(-4) = 0 …………………(i)

And f(2) can be calculated as

$\begin{align}

& f\left( 2 \right)={{\left( 2 \right)}^{2}}+2\left( 2 \right)-8 \\

& f(2)=4+4-8=8-8 \\

\end{align}$

Or

f(2) = 0 ……………..(ii)

Hence, from equation (i) and (ii) we get,

f(-4) = f(2) = 0

So, it is verified all three points for Rolle’s Theorem.

Now, for ${{f}^{'}}\left( c \right)$ , we have

$f\left( x \right)={{x}^{2}}+2x-8$

Differentiating f(x) w.r.t ‘x’, we get

${{f}^{'}}\left( x \right)=2x+1...........\left( iii \right)$

Where we know $\dfrac{d}{dx}\left( {{x}^{n}} \right)=n{{x}^{n-1}}$

Now, for equation (iii), we have

${{f}^{'}}\left( c \right)=2c+1$

Equating above equation to ‘0’, we can get value of ‘c’ as

$\begin{align}

& {{f}^{'}}\left( c \right)=0 \\

& \Rightarrow 2c+1=0 \\

& c=\dfrac{-1}{2} \\

\end{align}$

Where $c\in \left[ -4,2 \right]$

Hence, the given function $f\left( x \right)={{x}^{2}}+2x-8$ with $x\in \left( -4,2 \right)$ is satisfying Rolle’s theorem as

f(x) is continuous and differentiable in [-4,2] and (-4,2) respectively and f(-4) = f(2) = 0, then

${{f}^{'}}\left( \dfrac{-1}{2} \right)=0$ where $\dfrac{-1}{2}$ is lying between [-4,2].

Note: One can get confused between Rolle’s Theorem and mean value theorem. So, be careful with these kinds of questions and be clear with both the theorems.

One can factorize the given relation as, (x+4) (x-2) where x = -4 and x = 2 are roots of the given equation. Hence, we can directly write f(-4) = f(2) = 0 without putting values of x to a given function.

Recently Updated Pages

Calculate the entropy change involved in the conversion class 11 chemistry JEE_Main

The law formulated by Dr Nernst is A First law of thermodynamics class 11 chemistry JEE_Main

For the reaction at rm0rm0rmC and normal pressure A class 11 chemistry JEE_Main

An engine operating between rm15rm0rm0rmCand rm2rm5rm0rmC class 11 chemistry JEE_Main

For the reaction rm2Clg to rmCrmlrm2rmg the signs of class 11 chemistry JEE_Main

The enthalpy change for the transition of liquid water class 11 chemistry JEE_Main

Trending doubts

Difference Between Plant Cell and Animal Cell

Write an application to the principal requesting five class 10 english CBSE

Ray optics is valid when characteristic dimensions class 12 physics CBSE

Give 10 examples for herbs , shrubs , climbers , creepers

Write the 6 fundamental rights of India and explain in detail

Write a letter to the principal requesting him to grant class 10 english CBSE

List out three methods of soil conservation

Fill in the blanks A 1 lakh ten thousand B 1 million class 9 maths CBSE

Write a letter to the Principal of your school to plead class 10 english CBSE