Does a continuous function $f$ exist where:
$f$ is continuous,
$f$ is known to be bounded with a codomain of $\left[0,1\right]$
$f''$ (second derivative) is unbounded with a codomain of $(-\infty,+\infty)$
[Math] Unbounded second derivative of continuous bounded function
calculusreal-analysis
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Best Answer
Presumably the answer is yes: a function can be bounded but wiggle very vigorously.
So let's try to make one. Maybe we should start with $\cos(x)$, that's between $-1$ and $1$, easily fixed by adding $1$ and dividing by $2$. Unfortunately all derivatives are bounded. So give it a little more vigor by say looking at $f$ where $$f(x)=\frac{\cos(x^2) +1}{2}$$
We have $f'(x)=-x\sin(x^2)$ and $$f''(x)=-(2x^2\cos(x^2) +\sin(x^2))$$.
Now show that we can make $f''(x)$ arbitrarily large positive or negative. Then the Intermediate Value Theorem will show that the range of $f''(x)$ is the one you were looking for.
To make $f''(x)$ very large positive, just choose $x$ large such that $\cos(x^2)=-1$. To make $f''(x)$ very large negative, choose $x$ large such that $\cos(x^2)=1$. Each is easy to arrange.