In $[0, 1]$ Lagrange's mean value theorem is $ NOT$ applicable to
$f(x) = \left\{ {\begin{array}{*{20}{c}}
{\frac{1}{2} - x,\,\,\,\,\,\,\,\,\,x < \frac{1}{2}} \\
{{{\left( {\frac{1}{2} - x} \right)}^2},\,x \geqslant \frac{1}{2}}
\end{array}} \right.$
$f(x) = \left\{ {\begin{array}{*{20}{c}}
{\frac{{\sin x}}{x}\,\,x \ne 0} \\
{1,\,\,\,\,\,\,\,\,x = \frac{1}{2}}
\end{array}} \right.$
$f(x) = x|x|$
$f(x) = |x|$
If from mean value theorem, $f'({x_1}) = {{f(b) - f(a)} \over {b - a}}$, then
Verify Mean Value Theorem, if $f(x)=x^{2}-4 x-3$ in the interval $[a, b],$ where $a=1$ and $b=4$
Let $f$ be any function continuous on $[\mathrm{a}, \mathrm{b}]$ and twice differentiable on $(a, b) .$ If for all $x \in(a, b)$ $f^{\prime}(\mathrm{x})>0$ and $f^{\prime \prime}(\mathrm{x})<0,$ then for any $\mathrm{c} \in(\mathrm{a}, \mathrm{b})$ $\frac{f(\mathrm{c})-f(\mathrm{a})}{f(\mathrm{b})-f(\mathrm{c})}$ is greater than
Verify Mean Value Theorem, if $f(x)=x^{3}-5 x^{2}-3 x$ in the interval $[a, b],$ where $a=1$ and $b=3 .$ Find all $c \in(1,3)$ for which $f^{\prime}(c)=0$
If for $f(x) = 2x - {x^2}$, Lagrange’s theorem satisfies in $[0, 1]$, then the value of $c \in [0,\,1]$ is