If $f(a)=0$, $f^{(n)}(a)$ exists, then $F(x)=[f(x)]^n$ satisfies $F^{(k)}(a)=0,k=0,1,\cdots,n-1$.
This is easy to verify. But...
If $f^{(n)}(a)$ does not exist, could we find an example such that the conlcusion does not hold?
2026-03-25 17:44:35.1774460675
If $f(a)=0$, $f^{(n)}(a)$ exists, then $F(x)=[f(x)]^n$ satisfies $F^{(k)}(a)=0,k=0,1,\cdots,n-1$.
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