If the function $f(x) = - 4{e^{\left( {\frac{{1 - x}}{2}} \right)}} + 1 + x + \frac{{{x^2}}}{2} + \frac{{{x^3}}}{3}$ and $g(x)=f^{-1}(x) \,;$ then the value of $g'(-\frac{7}{6})$ equals
$\frac{1}{5}$
$- \frac{1}{5}$
$\frac{6}{7}$
$ -\frac{6}{7}$
In the mean value theorem, $f(b) - f(a) = (b - a)f'(c) $ if $a = 4$, $b = 9$ and $f(x) = \sqrt x $ then the value of $c$ is
From mean value theorem $f(b) - f(a) = $ $(b - a)f'({x_1});$ $a < {x_1} < b$ if $f(x) = {1 \over x}$, then ${x_1} = $
In which of the following functions is Rolle's theorem applicable ?
The function $f(x) = {x^3} - 6{x^2} + ax + b$ satisfy the conditions of Rolle's theorem in $[1, 3]. $ The values of $a $ and $ b $ are
If the function $f(x) = {x^3} - 6{x^2} + ax + b$ satisfies Rolle’s theorem in the interval $[1,\,3]$ and $f'\left( {{{2\sqrt 3 + 1} \over {\sqrt 3 }}} \right) = 0$, then $a =$ ..............