Verify Rolle’s Theorem for the function
$f\left( x \right)=\log \left\{ \dfrac{{{x}^{2}}+ab}{\left( a+b \right)x} \right\}$
In the interval [a, b] where, $0\notin \left[ a,b \right]$

Hint: Use the basic properties of logarithm and simplify f (x). Differentiate it to get ${{f}^{'}}\left( x \right)$ put ${{f}^{'}}\left( c \right)=0,x=c\Rightarrow {{f}^{'}}\left( x \right)$ and find the relation connect a, b, and c.

Complete step-by-step answer:
Rolle’s Theorem states that any real values differentiable function that attain equal value at two distinct points have at least one stationary point somewhere between them that is a point where the first derivative is zero. Here we have been given the function,
$f\left( x \right)=\log \left\{ \dfrac{{{x}^{2}}+ab}{\left( a+b \right)x} \right\}$
We know the basic logarithmic identities that,
$\begin{align}
  & \log \left( \dfrac{a}{b} \right)=\log a-\log b \\
 & \log \left( ab \right)=\log a+\log b \\
\end{align}$
Now apply all these identities in f (x)
$\begin{align}
  & f\left( x \right)=\log \left( \dfrac{{{x}^{2}}+ab}{\left( a+b \right)x} \right)...................\left( i \right) \\
 & =\log \left( {{x}^{{}}}+ab \right)-\log \left[ \left( a+b \right)x \right] \\
 & =\log \left( {{x}^{2}}+ab \right)-\left[ \log \left( a+b \right)+\log x \right] \\
 & f\left( x \right)=\log \left( {{x}^{2}}+ab \right)-\log \left( a+b \right)-\log x....................\left( ii \right) \\
\end{align}$
Now this is a continuous function of x in [a, b]. Now let us differentiate equation (ii) with respect to x, we get
$\begin{align}
  & f\left( x \right)=\log \left( {{x}^{2}}+ab \right)-\log \left( a+b \right)-\log \left( x \right) \\
 & \dfrac{d}{dx}\log x=\dfrac{1}{x} \\
 & {{f}^{'}}\left( x \right)=\dfrac{2x}{{{x}^{2}}+ab}-o-\dfrac{1}{x} \\
 & \therefore {{f}^{'}}\left( x \right)=\dfrac{2x}{{{x}^{2}}+ab}-\dfrac{1}{x}........................\left( iii \right) \\
\end{align}$
Since $x\in \left[ a,b \right]\Rightarrow x\ne 0$
Thus f (x) is differential in [a, b]. Now put x = a in equation (i)
$\begin{align}
  & f\left( a \right)=\log \left( \dfrac{{{a}^{2}}+ab}{\left( a+b \right)a} \right)=\log \left( \dfrac{{{a}^{2}}+ab}{{{a}^{2}}+ab} \right)=\log 1=0 \\
 & f\left( b \right)=\log \left( \dfrac{{{b}^{2}}+ab}{\left( a+b \right)b} \right)=\log \left( \dfrac{{{b}^{2}}+ab}{ab+{{b}^{2}}} \right)=\log 1=0 \\
\end{align}$
Thus f (x) satisfies all the condition of Rolle’s Theorem and therefor their exist at least one point x = c in (a, b) such that ${{f}^{'}}\left( c \right)=0$. Now put x = c in equation (iii)
$\dfrac{2c}{{{c}^{2}}+ab}-\dfrac{1}{c}=0$
Now let us simplify the above expression.
$\begin{align}
  & \dfrac{2{{c}^{2}}-\left( {{c}^{2}}+ab \right)}{c\left( {{c}^{2}}+ab \right)}=0 \\
 & \Rightarrow 2{{c}^{2}}-\left( {{c}^{2}}+ab \right)=0 \\
 & 2{{c}^{2}}-{{c}^{2}}-ab=0 \\
 & \therefore {{c}^{2}}=ab\Rightarrow c=1\sqrt{ab} \\
\end{align}$
Out of these two values $c=\sqrt{ab}$ , lies between a and b. Thus Rolle’s Theorem is verified in the interval [a, b]

Note: Rolle’s Theorem is a special case of mean value theorem. Now Rolle’s Theorem is different from Lagrange’s mean Value theorem. If a function is continuous in [a, b] and differentiate at open interval (a, b), then x = c exist in (a, b) such that ${{f}^{'}}\left( c \right)=0$ Remember this point in Rolle’s Theorem.