Question

Let $f\left(x\right)=x^2-2x+1,$ then the value of $c$ that satisfy the $LMVT$ for the function on the interval $[0,1]$ is

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

Answer 1

The correct option is A $\frac{1}{2}$

$f\left(x\right)$ is polynomial function so it is continuous and differentiable function on the interval $[0,1]$
By $LMVT,$ we get
$f^{\prime }\left(c\right)=\frac{f\left(1\right)-f\left(0\right)}{1-0}$
$\Rightarrow 2c-2=-1$
$\therefore c=\frac{1}{2}$