Question

For the function $f(x) = x^2 - 2x + 1$ on the interval $[0,2]$, what is the value of $c$ that satisfies Rolle's Theorem?

• A. $c=0$
• B. $c=1$
• C. $c=2$
• D. $c=-2$

Answer 1

Answer: (B)

$c=1$

Answer 2

To find the value of $c$ that satisfies Rolle's Theorem for the function $f(x) = x^2 - 2x + 1$ on the interval $[0, 2]$, we need to:

1. Verify conditions for Rolle's Theorem:

   • $f(x)$ must be continuous on $[0, 2]$.
   • $f(x)$ must be differentiable on $(0, 2)$.
   • $f(0) = f(2)$.

   Since $f(x)$ is a polynomial, it is continuous and differentiable everywhere. $f(0) = (0)^2 - 2(0) + 1 = 1$ $f(2) = (2)^2 - 2(2) + 1 = 1$ Thus, $f(0) = f(2) = 1$, satisfying all conditions.

2. Find the value of c: $f'(x) = 2x - 2$ Set $f'(c) = 0$: $2c - 2 = 0$ $2c = 2$ $c = 1$

3. Verify if c is in the interval (0, 2): Since $0 < 1 < 2$, $c=1$ lies within the interval.

Therefore, the value of $c$ that satisfies Rolle's Theorem is 1.