• 2022-05-31
    函数\( f\left( x \right) = x\sqrt {3 - x} \)在区间\( \left[ {0,3} \right] \)上满足罗尔定理的数值\( \xi \)= ______ .
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    内容

    • 0

      多项式\( f\left( x \right) = {x^3} - 3x + a \)在\( \left[ {0,1} \right] \)上有两个零点( )

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      已知函数\( f(x) \)在区间\( \left[ {a,b} \right] \)上连续,则由\( y = f(x),\;x = a,\;x = b,\;x \)轴围成的平面图形面积为( )。 A: \( \int_a^b {f(x)dx} \) B: \( \left| {\int_a^b {f(x)dx} } \right| \) C: \( \int_a^b {\left| {f(x)} \right|dx} \) D: \( f'(\xi )(b - a) \)

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      函数\( y = x \)在区间\( \left[ { - 1,1} \right] \)上满足罗尔定理( )

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      若\({y_1}\left( x \right), {y_2}\left( x \right)\)都是\(y' + P\left( x \right)y = Q\left( x \right)\)的特解,且 \({y_1}\left( x \right), {y_2}\left( x \right)\) 线性无关,则通解可表为\(y\left( x \right) = {y_1}\left( x \right) + C\left[ { { y_1}\left( x \right) - {y_2}\left( x \right)} \right]\)。

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      求函数$y = \root 3 \of {x + \sqrt x } $的导数$y' = $( ) A: ${{1 + 2\sqrt x } \over {\root 3 \of {{{\left( {x + \sqrt x } \right)}^2}} }}$ B: $ {{1 + 2\sqrt x } \over {6\root 3 \of {{{\left( {x + \sqrt x } \right)}^2}} }}$ C: $ {{1 + 2\sqrt x } \over {6\sqrt x \cdot \root 3 \of {{{\left( {x + \sqrt x } \right)}^2}} }}$ D: $ {{1 + 2\sqrt x } \over {\sqrt x \cdot \root 3 \of {{{\left( {x + \sqrt x } \right)}^2}} }}$