If f(x) is a twice differentiable polynomial | vedclass

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(B) Let $g(x) = f(x) - x^2$.Since $f(1) = 1, f(2) = 4, f(3) = 9$,we have $g(1) = 1 - 1^2 = 0$,$g(2) = 4 - 2^2 = 0$,and $g(3) = 9 - 3^2 = 0$.Thus,$g(x)

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There exists at least one $x \in (1, 3)$ such that $f''(x) = 2$

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(B) Let $g(x) = f(x) - x^2$.Since $f(1) = 1, f(2) = 4, f(3) = 9$,we have $g(1) = 1 - 1^2 = 0$,$g(2) = 4 - 2^2 = 0$,and $g(3) = 9 - 3^2 = 0$.Thus,$g(x)$ has at least $3$ real roots at $x = 1, 2, 3$.By Rolle's Theorem,$g'(x)$ has at least one root in $(1, 2)$ and at least one root in $(2, 3)$,meaning $g'(x)$ has at least $2$ roots in $(1, 3)$.Applying Rolle's Theorem again to $g'(x)$,$g''(x)$ must have at least $1$ root in $(1, 3)$.Since $g(x) = f(x) - x^2$,we have $g'(x) = f'(x) - 2x$ and $g''(x) = f''(x) - 2$.Since $g''(c) = 0$ for some $c \in (1, 3)$,it follows that $f''(c) - 2 = 0$,or $f''(c) = 2$ for at least one $c \in (1, 3)$.

If $f(x)$ is a twice differentiable polynomial function such that $f(1) = 1, f(2) = 4, f(3) = 9$,then:

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