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Hi was just wondering about this question, trying to show that the function is infinitely differentiable, my first attempt at this question was to differentiate the function several times at which point the function differentiated would be a constant but I am not sure if this shows that the function is infinitely differentiable when it reaches the point at which the differential is 0.

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    $\begingroup$ Just take three derivatives. Done! $\endgroup$ – David G. Stork Oct 2 '19 at 22:33
  • $\begingroup$ how does that imply that its infinitely differentiable can you please elaborate @DavidG.Stork $\endgroup$ – John Oct 2 '19 at 22:40
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    $\begingroup$ Because once you get to the third derivative, all are $0$. ALL derivatives exist. So the function is $C^\infty$ which means continuous. $\endgroup$ – David G. Stork Oct 2 '19 at 22:41
  • $\begingroup$ ok thanks for the confirmation and the explanation @DavidG.Stork $\endgroup$ – John Oct 2 '19 at 22:42
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In general, a polynomial of degree $n$ $$p(x)=a_0+a_1x+a_2x^2+\dots+a_nx^n$$ defined over $\mathbb R$ is always infinitely differentiable. This is because by the power rule (which I assume you've learnt before), $$\frac{d}{dx}a_kx^k=ka_kx^{k-1}$$ for every $0\leq k\leq n$, so by the linearity of the derivative, $p'(x)$ exists and is equal to $$\frac{d}{dx}p(x)=a_1+2a_2x+3a_3x^2+\dots+na_nx^{n-1}.$$ Inductively this establishes the existence of all the derivatives for polynomials of all degrees.

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