# When is $f''(x)$ undefined but $x$ not an inflection point of $f$?

This is supposed to be for bullet 5.2 in this question: A rigorous yet intuitive summary of inflection and critical points for beginning calculus?

• There are examples for undulation points, where $$f: A \to B, A,B \subseteq \mathbb R, f''(x)=0$$ but $$(x,f(x))$$ is not an inflection point of $$f$$ like $$f(t) = \frac{t^4}{12},f''(t)=t^2$$ (If $f''(x)=0$ but is not an inflection point, what is it called?)

• There are examples where $$f''(x)$$ is undefined but $$(x,f(x))$$ is an inflection point like $$f'(t)=|t|$$ (An inflection point where the second derivative doesn't exist?)

I'm missing something very obvious, but what's an example where $$f''(x)$$ is undefined, but $$(x,f(x))$$ is not an inflection point of $$f$$?

After some thought, I have come up with some examples. Are these correct?

First example: This example is one where $$f$$ is defined but not continuous at $$x$$.

• $$f: \mathbb R \to \mathbb R, f(t)=\text{sign}(t), x=0$$. $$f'(0)$$ is undefined and thus so is $$f''(0)$$. However $$f''(t)=0$$ for $$t \ne 0$$.

• In case someone asks further 'Okay, what if we assumed $$f$$ is continuous? Then $$x$$ would have to be an inflection point right?', then we proceed:

Second example: This example is one where $$f$$ is continuous but not differentiable at $$x$$.

• A sharp turn like $$g(t)=|t|$$ but always increasing at an increasing rate but at different rates $$f: \mathbb R \to \mathbb R, f(t)= t 1_{t \le 0} + 3t 1_{t \ge 0}$$. This gives us $$f': \mathbb R \setminus \{0\} \to \mathbb R, f'(t) = \text{sign}(t)+2$$. $$f'(0)$$ is undefined and thus so is $$f''(0)$$. However $$f''(t)>0$$ for $$t \ne 0$$.

• In case someone asks further 'Okay, what if we assumed $$f$$ is differentiable? Then $$x$$ would have to be an inflection point right?', then we proceed:

Third example: This example is one where $$f$$ is differentiable at $$x$$.

• $$f': \mathbb R \to \mathbb R, f'(t)= t 1_{t \le 0} + 3t 1_{t \ge 0}$$.
• Case 3 seems like the best answer here, since $f’’(0)$ isn’t defined in the first two cases. I would consider case 3 a good answer to this. – David M. Feb 24 at 13:11
• @DavidM. $f''(0)$ is supposed to be undefined in all cases. What do you mean? – Mitjackson Feb 24 at 16:30
• Oops I completely misread. What exactly is your question? – David M. Feb 24 at 16:40
• @DavidM. Edited: This is supposed to be for bullet 5.2 in this question: A rigorous yet intuitive summary of inflection and critical points for beginning calculus? – Mitjackson Feb 28 at 5:38

I can’t tell what you're asking, so I’ll answer the question in the title. Or, more precisely, I’ll give an example of an $$f$$, differentiable on the whole real line, with the property you describe in the title. I can’t tell what you’re doing by listing all those cases.
Take $$f$$ to be $$x^2$$ when $$x$$ is negative and $$x^4$$ when $$x$$ is non-negative.
Then $$f’’(0)$$ is undefined, and moreover $$f$$ does not have an inflection point at the origin.
• Thanks! I was giving several answers with different conditions. One might say after the first answer $f(t) = sign(t)$, something like 'Okay, what if we assumed $f$ is continuous? Then $x$ would have to be an inflection point right?' Then comes the second answer. Does that clarify things? – Mitjackson Feb 28 at 6:06