Let f :[0,1] rightarrow R be a twice differen | vedclass

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(C) Define $g(x) = f(x) - (2x + 3)$.Given $f(0)=3$,so $g(0) = f(0) - (2(0) + 3) = 3 - 3 = 0$.Given $f(1)=5$,so $g(1) = f(1) - (2(1) + 3) = 5 - 5 = 0$.

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$2$

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(C) Define $g(x) = f(x) - (2x + 3)$.Given $f(0)=3$,so $g(0) = f(0) - (2(0) + 3) = 3 - 3 = 0$.Given $f(1)=5$,so $g(1) = f(1) - (2(1) + 3) = 5 - 5 = 0$.Since the line $y=2x+3$ intersects $f(x)$ at two distinct points in $(0,1)$,let these points be $x_1$ and $x_2$ where $0 Thus,$g(x_1) = 0$ and $g(x_2) = 0$.We have $g(0)=0, g(x_1)=0, g(x_2)=0, g(1)=0$.By Rolle's Theorem,$g^{\prime}(x)$ must have at least one root in each of the intervals $(0, x_1)$,$(x_1, x_2)$,and $(x_2, 1)$.Let these roots be $c_1, c_2, c_3$ such that $0 Now,applying Rolle's Theorem to $g^{\prime}(x)$ on the intervals $(c_1, c_2)$ and $(c_2, c_3)$:$g^{\prime\prime}(x)$ must have at least one root in $(c_1, c_2)$ and at least one root in $(c_2, c_3)$.Therefore,$g^{\prime\prime}(x) = f^{\prime\prime}(x)$ has at least $2$ roots in $(0,1)$.

Let $f :[0,1] \rightarrow R$ be a twice differentiable function in $(0,1)$ such that $f(0)=3$ and $f(1)=5$. If the line $y=2x+3$ intersects the graph of $f$ at only two distinct points in $(0,1)$,then the least number of points $x \in(0,1)$,at which $f^{\prime\prime}(x)=0$,is $......$

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