Question

If $f\left(x\right)=x(x-2)(x-4),1\le x\le 4,$ then a number satisfying the conditions of the mean value theorem is

• A. $1$
• B. $\frac{5}{2}$
• C. $3$
• D. $\frac{7}{2}$

Answer 1

The correct option is C $3$

From lagrange theorem, we can say that
$\frac{f\left(4\right)-f\left(1\right)}{4-1}=f^{\prime }\left(c\right)=-1$
Now,

$f^{\prime }\left(c\right)=(c-2)(c-a)+c(c-4)+c(c-2)=-1$

Solving the above equation, we get

$c=1$ or $3$

Both values lie in $[1,4]$, but according to Lagrange theorem, $c$ must lie in open intervals i.e. $c\in (1,4)\equiv c=3$, only

Hence, option C.