Let f be any continuous function on [0,2] and | vedclass

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(B) Given $f(0)=0, f(1)=1$,and $f(2)=2$. Define a function $h(x) = f(x) - x$. Then $h(0) = f(0) - 0 = 0$,$h(1) = f(1) - 1 = 0$,and $h(2) = f(2) - 2 =

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$f^{\prime \prime}(x)=0$ for some $x \in(0,2)$

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(B) Given $f(0)=0, f(1)=1$,and $f(2)=2$. Define a function $h(x) = f(x) - x$. Then $h(0) = f(0) - 0 = 0$,$h(1) = f(1) - 1 = 0$,and $h(2) = f(2) - 2 = 0$. Since $h(x)$ is continuous on $[0,1]$ and $[1,2]$ and differentiable on $(0,1)$ and $(1,2)$,by Rolle's Theorem,there exists $c_1 \in (0,1)$ such that $h^{\prime}(c_1) = 0$ and $c_2 \in (1,2)$ such that $h^{\prime}(c_2) = 0$. Now,$h^{\prime}(x) = f^{\prime}(x) - 1$. Since $h^{\prime}(c_1) = 0$ and $h^{\prime}(c_2) = 0$,and $h^{\prime}(x)$ is continuous on $[c_1, c_2]$ and differentiable on $(c_1, c_2)$,by Rolle's Theorem applied to $h^{\prime}(x)$,there exists at least one $c \in (c_1, c_2) \subset (0,2)$ such that $h^{\prime \prime}(c) = 0$. Since $h^{\prime \prime}(x) = f^{\prime \prime}(x)$,we have $f^{\prime \prime}(c) = 0$ for some $c \in (0,2)$.

Let $f$ be any continuous function on $[0,2]$ and twice differentiable on $(0,2)$. If $f(0)=0, f(1)=1$ and $f(2)=2$,then

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