Let f:(a, b) rightarrow R be a twice differen | vedclass

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(A) Given $f(x) = \int_{a}^{x} g(t) \, dt$. By the Fundamental Theorem of Calculus,$f'(x) = g(x)$.Since $f(x) = 0$ has $5$ distinct roots in $(a, b)$,

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seven roots in $(a, b)$

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(A) Given $f(x) = \int_{a}^{x} g(t) \, dt$. By the Fundamental Theorem of Calculus,$f'(x) = g(x)$.Since $f(x) = 0$ has $5$ distinct roots in $(a, b)$,by Rolle's Theorem,$f'(x) = 0$ must have at least $5 - 1 = 4$ distinct roots in $(a, b)$.Thus,$g(x) = 0$ has at least $4$ distinct roots in $(a, b)$.Now,by Rolle's Theorem applied to $g(x)$,since $g(x) = 0$ has at least $4$ distinct roots,$g'(x) = 0$ must have at least $4 - 1 = 3$ distinct roots in $(a, b)$.The equation $g(x) g'(x) = 0$ is satisfied if $g(x) = 0$ or $g'(x) = 0$.Since $g(x) = 0$ has at least $4$ roots and $g'(x) = 0$ has at least $3$ roots,the total number of distinct roots for $g(x) g'(x) = 0$ is at least $4 + 3 = 7$ roots in $(a, b)$.

Let $f:(a, b) \rightarrow R$ be a twice differentiable function such that $f(x) = \int_{a}^{x} g(t) \, dt$ for a differentiable function $g(x)$. If $f(x) = 0$ has exactly five distinct roots in $(a, b)$,then $g(x) g'(x) = 0$ has at least:

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