Question

The value of $c$ in $\left(0,2\right)$ satisfying the mean value theorem for the function $f\left(x\right)=x{(x-1)}^2$, $x\in [0,2]=$?

• A. $\frac{3}{4}$
• B. $\frac{4}{3}$
• C. $\frac{1}{3}$
• D. $\frac{2}{3}$
• E. $\frac{5}{3}$

Answer 1

The correct option is B

$\frac{4}{3}$

Explanation for the correct option:

Find the value of $c$

A function $f\left(x\right)=x{(x-1)}^2$ is given with the interval $[0,2]$.

Lagrange's mean value theorem states that, if a function $g$ is continuous on $[a,b]$ then there exists a $c$ value between $a$ and $b$ for which $f\text{'}\left(c\right)=\frac{f\left(b\right)-f\left(a\right)}{b-a}$.

Find the derivative of the given at $x=c$.

$$\begin{gathered} f\text{'}(x=c)={(c-1)}^2+2c(c-1) \\ \Rightarrow f\text{'}(x=c)=(c-1)(3c-1) \\ \Rightarrow f\text{'}(x=c)=3c^2-4c+1 \end{gathered}$$

Apply Lagrange's mean value theorem as follows,

$$\begin{gathered} 3c^2-4c+1=\frac{2{(2-1)}^2-0{(0-1)}^2}{2-0} \\ \Rightarrow 3c^2-4c+1=\frac{2}{2} \\ \Rightarrow 3c^2-4c+1=1 \\ \Rightarrow c\left(3c-4\right)=0 \\ \Rightarrow c=0,\frac{4}{3} \end{gathered}$$

Since, $c=0$ does not lie between $0$ and $2$.

Therefore, the value of $c$ is $\frac{4}{3}$.

Hence, option $B$ is correct .