If f(x) and g(x) are twice differentiable fun | vedclass

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(B) Given that $f^{\prime \prime}(x)=g^{\prime \prime}(x)$.Integrating both sides with respect to $x$,we get $f^{\prime}(x)=g^{\prime}(x)+C_1$.Substit

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$-4$

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(B) Given that $f^{\prime \prime}(x)=g^{\prime \prime}(x)$.Integrating both sides with respect to $x$,we get $f^{\prime}(x)=g^{\prime}(x)+C_1$.Substituting $x=1$,we have $f^{\prime}(1)=g^{\prime}(1)+C_1$.Given $f^{\prime}(1)=4$ and $g^{\prime}(1)=6$,so $4=6+C_1$,which implies $C_1=-2$.Thus,$f^{\prime}(x)=g^{\prime}(x)-2$.Integrating again with respect to $x$,we get $f(x)=g(x)-2x+C_2$.Substituting $x=2$,we have $f(2)=g(2)-2(2)+C_2$.Given $f(2)=3$ and $g(2)=9$,so $3=9-4+C_2$,which implies $3=5+C_2$,so $C_2=-2$.Thus,$f(x)=g(x)-2x-2$.Rearranging the terms,we get $f(x)-g(x)=-2x-2$.For $x=1$,$f(1)-g(1)=-2(1)-2=-4$.

If $f(x)$ and $g(x)$ are twice differentiable functions on $(0,3)$ satisfying $f^{\prime \prime}(x)=g^{\prime \prime}(x)$,$f^{\prime}(1)=4$,$g^{\prime}(1)=6$,$f(2)=3$,and $g(2)=9$,then $f(1)-g(1)$ is

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