Sign up ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Find k such that the function $f(x)=|x|^3$ is $C^{k}$ but not $C^{k+1}$

I am SO lost...any help would be appreciated

share|cite|improve this question

1 Answer 1

up vote 7 down vote accepted

Well, $$ f(x)= \begin{cases} x^3 &\text{if $x \geq 0$}\\ -x^3 &\text{if $x < 0$}. \end{cases} $$ So you are gluing two function at $x=0$: the derivatives of the two pieces are $3x^2$, $6x$, $6$, on the right, and $-3x^2$, $-6x$, $-6$ on the left. Higher derivatives are identically zero. Now the question is: how manu derivatives can you glue at zero, so that the gluing is continuous? The answer is not very difficult.

share|cite|improve this answer
Wow...I had a major brain fart there! Thanks for the clarification, I am embarrassed. – xc92 Apr 27 '14 at 11:37

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.