Verified answer

Try something like f(x) = x^6 sin(1/x) for nonzero x, and f(0) = 0.

f '(x) = 6x^5 sin(1/x) - x^4 cos(1/x) for nonzero x, and

f '(0) = 0 by the definition of the derivative and the Squeeze Theorem.

Details:

f '(0) = lim(h→0) [f(0+h) - f(0)]/h

.......= lim(h→0) [h^6 sin(1/h) - 0]/h

.......= lim(h→0) h^5 sin(1/h)

.......= 0, since -|h^5| ≤ h^5 sin(1/h) ≤ h^5 for all nonzero h, and lim(h→0) ±|h^5| = 0.

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Similarly,

f ''(x) = 30x^4 sin(1/x) - 10x^3 cos(1/x) + x^2 sin(1/x) for nonzero x, and f ''(0) = 0 (as before).

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Finally, f '''(x) = 120x^3 sin(1/x) - 60x^2 cos(1/x) - 8x sin(1/x) - cos(1/x), while f '''(0) = 0.

However, lim(x→0) f '''(x) does not exist, due to the cos(1/x) term.

So, f ''' is discontinuous at x = 0, although f '''(0) is defined.

I hope this helps!

sure. The consistent functionality 0 satisfies this supplies. EDIT: of direction, you explicitly reported the codomain replaced into the useful real numbers, so 0 is out of the question. provide me a 2nd to think of. EDIT: ok, after thinking approximately it for an afternoon, attempting numerous strategies, attempting the two to construct this way of functionality and teach that one does not exist, and asking my professor (who's additionally at present thinking approximately it), I surrender. The "why?" question on the top has a tendency to signify you be attentive to the respond is that it does not exist, so which you tell me: why no longer? EDIT: If we are speaking properties this functionality has, if it exists, right here is what i found out: a million) it relatively is totally discontinuous (as has been reported) 2a) For any sequence x(n) converging to a pair y interior the irrationals, the sequence f(x(n)) converges to 0 2b) comparable ingredient, different than change the rationals for irrationals 3) whether the situation has some sort of "invariance" below addition with the aid of a rational (in that f(x + q)f(y + q) <= |x - y|), the functionality, with the aid of 2a), can not probably be invariant below addition with the aid of a rational 4) If the functionality exists on some neighbourhood, then we can build a functionality such that it satisfies the situation on all of R. 5) I *think of* i can, for each and all the reliable it relatively is going to do, build this way of functionality that works for rational x and irrational, yet algebraic, y. Or, greater precisely, irrational, yet no longer a Liouville huge type. I nevertheless could desire to employer up the main factors however.