Question

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

• 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 correct option:
Given function is $f\left(x\right)=x{(x-1)}^2,x\in \left[0,2\right]$
Mean Value Theorem : The Mean Value Theorem states that if a function is continuous on the closed interval $[a,b]$ and differentiable on the open interval $(a,b)$, then there exists a point $c$ in the interval $(a,b)$ such that $f\text{'}\left(c\right)$ is equal to the function's average rate of change over $[a,b]$ i.e. $\mathbf{}\mathbf{f}\mathbf{\text{'}}\mathbf{\left(}\mathbf{c}\mathbf{\right)}\mathbf{=}\frac{\mathbf{f}\mathbf{\left(}\mathbf{b}\mathbf{\right)}\mathbf{-}\mathbf{f}\mathbf{\left(}\mathbf{a}\mathbf{\right)}}{\mathbf{b}\mathbf{-}\mathbf{a}}$.
$$\begin{gathered} f\left(x\right)=x{(x-1)}^2 \\ \Rightarrow f\left(x\right)=x\left(x^2-2x+1\right)\left[\because {\left(a-b\right)}^2=a^2-2\mathrm{ab}+b^2\right] \\ \Rightarrow f\left(x\right)=x^3-2x^2+x \\ \Rightarrow f\left(2\right)=2^3-2·2^2+2=2 \\ \Rightarrow f\left(0\right)=0^3-2·0^2+0=0 \\ f\text{'}\left(x\right)=3x^2-4x+1 \end{gathered}$$
According to theorem
$$\begin{gathered} \Rightarrow f\text{'}\left(c\right)=\frac{f\left(b\right)-f\left(a\right)}{b-a} \\ \Rightarrow 3c^2-4c+1=\frac{f\left(2\right)-f\left(0\right)}{2-0} \\ \Rightarrow 3c^2-4c+1=\frac{2}{2} \\ \Rightarrow 3c^2-4c=1-1 \\ \Rightarrow 3c^2=4c \\ \Rightarrow c=\frac{4}{3} \end{gathered}$$

Hence option(B) is correct