Is anyone able to provide a(n) example(s) of strictly positive, increasing, concave function where the second derivative becomes larger (in absolute value) as $x$ increases?
Or, if this is not possible, can someone provide some intuition or a proof/link to explain?
Thanks
Not on all of $(-\infty,\infty)$, no. In fact, if $f$ is increasing and concave on $(-\infty,\infty)$ then it must satisfy $\lim_{x\to-\infty}f(x)=-\infty$.
To see this, note that since $f$ is increasing, its derivative exists at some $x_0\in(-\infty,\infty)$. Since $f'(x_0)>0$, the tangent line $T(x)$ to $f$ at $x_0$ has positive slope, so that $T(x)\to-\infty$ as $x\to-\infty$. However due to concavity of $f$ we have $f(x)\leq T(x)$ for all $x\in(-\infty,\infty)$, and hence $f(x)\to -\infty$ as $x\to-\infty$ as claimed.
It's also impossible to find an example on a positive ray, although for different reasons. Without loss of generality we may suppose $f:(0,\infty)\to(-\infty,\infty)$ is concave and that $f''$ exists with $|f''(x)|$ increasing. Since $f''(x)<0$, that means $f''$ is decreasing. In particular, there are $a,\delta>0$ such that $f'(x)<-\delta$ for all $x\geq a$. Note that there is $C\in(-\infty,\infty)$ such that $f'(x)=C+\int_a^xf''(t)\;dt\leq C+\delta a-\delta x\to -\infty$ as $x\to\infty$. Hence $f$ is eventually negative.
If you want $f$ defined on an interval, that means the interval must be bounded. This is quite easy to do. For instance let $f:(0,1)\to(0,\infty)$ be defined by $f(x)=3x-x^3$.
