设f(x)={3x+a,x⩽02x2+1,0<x⩽1bx,x>1f(x)=\begin{cases}3x + a, &x\leqslant0\\2x^{2}+1, &0<x\leqslant1\\\dfrac{b}{x}, &x > 1\end{cases}f(x)=⎩⎨⎧3x+a,2x2+1,xb,x⩽00<x⩽1x>1在x=0x = 0x=0和x=1x = 1x=1处连续,求aaa,bbb的值.
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