Let $f ( x ) = x ^ { 2 } - 2 x - 2$ on $[ 0,2 ]$ . Then the set of all c in (0,2) guaranteed by Rolle's Theorem is

A) $\{ - 2 \}$
B) $\{ - 1 \}$
C) $\{ 0 \}$
D) $\{ 1 \}$
E) $\{ 2 \}$

Question

Let $f ( x ) = x ^ { 2 } - 2 x - 2$ on $[ 0,2 ]$ . Then the set of all c in (0,2) guaranteed by Rolle's Theorem is

A) $\{ - 2 \}$
B) $\{ - 1 \}$
C) $\{ 0 \}$
D) $\{ 1 \}$
E) $\{ 2 \}$

Quizplus · Accepted Answer

Rolle's Theorem states that if a function $f$ is continuous on a closed interval $[a, b]$, differentiable on the open interval $(a, b)$, and $f(a) = f(b)$, then there exists at least one $c$ in $(a, b)$ such that $f'(c) = 0$. For $f(x) = x^2 - 2x - 2$ on $[0, 2]$, we first check if $f(0) = f(2)$. Indeed, $f(0) = 0^2 - 2(0) - 2 = -2$ and $f(2) = 2^2 - 2(2) - 2 = -2$, so the condition $f(0) = f(2)$ is satisfied. Next, we find $f'(x) = 2x - 2$. Setting $f'(x) = 0$ gives $2x - 2 = 0$, which solves to $x = 1$. Therefore, $c = 1$ is the value in $(0, 2)$ that satisfies Rolle's Theorem.

Let f(x)=x2−2x−2f ( x ) = x ^ { 2 } - 2 x - 2f(x)=x2−2x−2

Let $f ( x ) = x ^ { 2 } - 2 x - 2$