Answer

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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.

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