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 verify the conditions and then find the value of $c$ such that $f'(c) = 0$.

1. Conditions for Rolle's Theorem:

   • $f(x)$ is continuous on $[0, 2]$.
   • $f(x)$ is 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.