Question

Verify Lagrange’s mean value theorem for the following function on the indicated interval. In each case find a point ‘c’ in the indicated interval as stated by the Lagrange’s mean value theorem: $f\left( x \right)=x{{\left( x+4 \right)}^{2}}$ on $\left[ 0,4 \right]$.

Answer 1

Hint: We first define Lagrange's mean value theorem. We take the values and the extremum points. We place the differentiation of the given function. We find a fixed point c for which the function satisfies the formula.

Complete step by step answer:
Lagrange’s mean value theorem tells us that if $f\left( x \right)$ be a real valued function of the real variable x defined in $a\le x\le b$ such that
(i) it is continuous in $a\le x\le b$
(ii) it’s differentiable in $a\le x\le b$;
then there is at least one value of $x=c$ between a and b $\left( a < c < b \right)$ such that $\dfrac{f\left( b \right)-f\left( a \right)}{b-a}={{f}^{'}}\left( c \right)$.
In our given problem we need to verify Lagrange’s mean value theorem for $f\left( x \right)=x{{\left( x+4 \right)}^{2}}$ on $\left[ 0,4 \right]$.
So, we find value for the formula $\dfrac{f\left( b \right)-f\left( a \right)}{b-a}={{f}^{'}}\left( c \right)$.
We assume the value c is in between a and b $\left( a < c < b \right)$. Here a and b are 0 and 4 respectively. We need to find the value of c.
$f\left( 0 \right)=0{{\left( 0+4 \right)}^{2}}=0$ and $f\left( 4 \right)=4{{\left( 4+4 \right)}^{2}}=256$.
We find the value of ${{f}^{'}}\left( x \right)$ by differentiating both sides of $f\left( x \right)=x{{\left( x+4 \right)}^{2}}$.
So, ${{f}^{'}}\left( x \right)={{\left( x+4 \right)}^{2}}+2x\left( x+4 \right)=\left( x+4 \right)\left( x+4+2x \right)=\left( x+4 \right)\left( 3x+4 \right)$.
Now we find value of ${{f}^{'}}\left( c \right)$.
 ${{f}^{'}}\left( c \right)=\left( c+4 \right)\left( 3c+4 \right)$.
We place the values in equation $\dfrac{f\left( b \right)-f\left( a \right)}{b-a}={{f}^{'}}\left( c \right)$ where $a=0,b=4$.
$\begin{align}
  & \dfrac{f\left( 4 \right)-f\left( 0 \right)}{4-0}={{f}^{'}}\left( c \right) \\
 & \Rightarrow \dfrac{256-0}{4}=\left( c+4 \right)\left( 3c+4 \right) \\
 & \Rightarrow \left( c+4 \right)\left( 3c+4 \right)=64 \\
 & \Rightarrow {{c}^{2}}+16c-48=0 \\
\end{align}$
We solve the quadratic equation of c to find possible values of c.
$\begin{align}
  & {{c}^{2}}+16c-48=0 \\
 & \Rightarrow c=\dfrac{-16\pm \sqrt{{{16}^{2}}+4\times 1\times 48}}{2} \\
 & \Rightarrow c=\dfrac{-16\pm 8\sqrt{7}}{2} \\
\end{align}$
Now the value of c has to be in between 0 and 4.
c has two values $c=\dfrac{-16-8\sqrt{7}}{2},\dfrac{-16+8\sqrt{7}}{2}$.
The value of $c=\dfrac{-16-8\sqrt{7}}{2}$ is less than 0. So, we don’t need this solution.
But the approximate value of $c=\dfrac{-16+8\sqrt{7}}{2}=4\sqrt{7}-8$ is $c=\text{2}\text{.583}$. This is in between 0 and 4.
This means there exists one value of $x=c=4\sqrt{7}-8$ between 0 and 4 $\left( 0<4\sqrt{7}-8<4 \right)$ such that $\dfrac{f\left( 4 \right)-f\left( 0 \right)}{4-0}={{f}^{'}}\left( 4\sqrt{7}-8 \right)$.

Therefore, $f\left( x \right)=x{{\left( x+4 \right)}^{2}}$ on $\left[ 0,4 \right]$ satisfies Lagrange’s mean value theorem.

Note: The main condition of the theorem is for c to be in between a and b. the root of the quadratic has to satisfy the condition otherwise it will never satisfy the theorem. That’s why we omitted the other root of the equation.