若f(x+3)=x2+6x+11f(x + 3)=x^2 + 6x + 11f(x+3)=x2+6x+11，求f(x)f(x)f(x).
若函数f(x)={k+x2,x≤0x+1−1x,x>0f(x)=\begin{cases}k + x^2, & x \leq 0\\\dfrac{\sqrt{x + 1}-1}{x}, & x > 0\end{cases}f(x)=⎩⎨⎧k+x2,xx+1−1,x≤0x>0在(−∞,+∞)(-\infty,+\infty)(−∞,+∞)内连续，求kkk的值.
已知函数y=3xy = 3^xy=3x，则dy=dy =dy=（ ）.
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