Let f, g:[-1,2] rightarrow R be continuous fu | vedclass

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(D) Define $H(x) = f(x) - 3g(x)$.Calculating values of $H(x)$ at given points:$H(-1) = f(-1) - 3g(-1) = 3 - 3(0) = 3$.$H(0) = f(0) - 3g(0) = 6 - 3(1)

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$(B, C)$

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(D) Define $H(x) = f(x) - 3g(x)$.Calculating values of $H(x)$ at given points:$H(-1) = f(-1) - 3g(-1) = 3 - 3(0) = 3$.$H(0) = f(0) - 3g(0) = 6 - 3(1) = 3$.$H(2) = f(2) - 3g(2) = 0 - 3(-1) = 3$.Since $H(-1) = H(0) = 3$,by Rolle's Theorem,there exists at least one $c_1 \in (-1, 0)$ such that $H^{\prime}(c_1) = 0$.Since $H^{\prime \prime}(x)$ never vanishes in $(-1, 0)$,$H^{\prime}(x)$ is strictly monotonic,meaning there is exactly one solution in $(-1, 0)$.Similarly,since $H(0) = H(2) = 3$,by Rolle's Theorem,there exists at least one $c_2 \in (0, 2)$ such that $H^{\prime}(c_2) = 0$.Since $H^{\prime \prime}(x)$ never vanishes in $(0, 2)$,$H^{\prime}(x)$ is strictly monotonic,meaning there is exactly one solution in $(0, 2)$.Thus,statements $(B)$ and $(C)$ are correct.

$x$	$x=-1, 0, 2$
$f(x)$	$3, 6, 0$
$g(x)$	$0, 1, -1$

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