The value of c in Lagrange's mean value theorem

Solution:

(d) $\frac{3}{2}$

We have

$f(x)=x(x-2)$

It can be rewritten as $f(x)=x^{2}-2 x$.

We know that a polynomial function is everywhere continuous and differentiable.

Since $f(x)$ is a polynomial, it is continuous on $[1,2]$ and differentiable on $(1,2)$.

Thus, $f(x)$ satisfies both the conditions of Lagrange's theorem on $[1,2]$.

 

So, there must exist at least one real number $c \in(1,2)$ such that

$f^{\prime}(c)=\frac{f(2)-f(1)}{2-1}=\frac{f(2)-f(1)}{1}$

Now, $f(x)=x^{2}-2 x$

$\Rightarrow f^{\prime}(x)=2 x-2$,

and $f(1)=-1, f(2)=0$

$\therefore f^{\prime}(x)=\frac{f(2)-f(1)}{2-1}$

$\Rightarrow f^{\prime}(x)=\frac{0+1}{1}$

$\Rightarrow 2 x-2=1$

 

$\Rightarrow x=\frac{3}{2}$

$\therefore c=\frac{3}{2} \in(1,2)$