Find 'c' of the mean value theorem, if $\displaystyle f(x)=x(x-1)(x-2);a=0, b=1/2$

We have $f\left( a \right) =f\left( 0 \right) =0$ and $\displaystyle f\left( b \right) =f\left( \frac { 1 }{ 2 } \right) =\frac { 3 }{ 8 }$

$\displaystyle \therefore \frac { f\left( b \right) -f\left( a \right) }{ b-a } =\frac { \frac { 3 }{ 8 } -0 }{ \frac { 1 }{ 2 } -0 } =\frac { 3 }{ 4 }$

Now

$f\left( x \right) ={ x }^{ 3 }-3{ x }^{ 2 }+2x\\ \Rightarrow f'\left( x \right) ={ 3x }^{ 2 }-6x+2\\ \Rightarrow f'\left( c \right) ={ 3c }^{ 2 }-6c+2$

Putting all these value in lagrange's mean value theorem

$\displaystyle \frac { f\left( b \right) -f\left( a \right) }{ b-a } =f'\left( c \right) ,\left( a<c,b \right)$

We get $\displaystyle \frac { 3 }{ 4 } ={ 3c }^{ 2 }-6c+2\Rightarrow c=1\pm \frac { \sqrt { 21 } }{ 6 }$

Hence $\displaystyle c=\frac { 1-\sqrt { 21 } }{ 6 }$ lies in the open interval $\displaystyle \left( 0,\frac { 1 }{ 2 } \right)$

Therefore it is the required value