Question

The value of c in Lagranges mean value theorem for $f\left(x\right)=x{(x-2)}^2$ in $[0,2]$ is

• A. $0$
• B. $2$
• C. $\frac{2}{3}$
• D. $\frac{3}{2}$

Answer 1

The correct option is C $\frac{2}{3}$

Lagrange's mean value theorem states that if $f\left(x\right)$ be continuous on $[a,b]$ and differentiable on $(a,b)$ then there exists some $c$ in $(a,b)$ such that $f^{{}^{\prime }}\left(c\right)=\frac{f\left(b\right)-f\left(a\right)}{b-a}$

Given $f\left(x\right)=x(x-2{)}^2=x^3-4x^2+4x$ and $[a,b]=[0,2]$

$⟹f^{{}^{\prime }}\left(x\right)=3x^2-8x+4$

Therefore, $f^{{}^{\prime }}\left(c\right)=\frac{(b^3-4b^2+4b)-(a^3-4a^2+4a)}{b-a}$

$⟹3c^2-8c+4=\frac{2^3-4(2{)}^2+4(2)-0}{2-0}$

$⟹3c^2-8c+4=2^2-4\left(2\right)+4$

$⟹3c^2-8c+4=0$

$⟹3c^2-6c-2c+4=0$

$⟹3c(c-2)-2(c-2)=0$

$⟹(3c-2)(c-2)=0$

$⟹c=2/3$or $⟹c=2$

Since, $c=2\notin (0,2)$. Therefore $c=2/3\in (0,2)$