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This is a problem from Apostol's Calculus (p. 209 Ex. 26 (c) & (d)). The problem is to find a function $f$ with a continuous second derivative f'' satisfying the following conditions:

(c) f''(x) > 0 \quad \text{for every } x, \qquad f'(0) = 1, \qquad f(x) \leq 100 \quad \text{for all } x > 0.

(d) f''(x) > 0 \quad \text{for every } x, \qquad f'(0) = 1, \qquad f(x) \leq 100 \quad \text{for all } x < 0.

So, for part (c) I do not think such a function can exist. My proof is that f''(x) > 0 for every x \implies f'(x) is increasing. Since f'(0) = 1 this means f'(x) > 1 for $x > 0$.

By the mean value theorem we have f(b) - f(0) = f'(c) (b) for some $c \in (0,b)$. Since f'(c) > 1 for all $c \in (0,b)$ we have $f(b) > b + f(0)$ for any $b > 0$. So, just choose $b > 100 + |f(0)|$ to obtain $f(b) > 100$ contradicting that $f(x) \leq 100$ for all $x > 0$.

Is this a sensible approach? I feel like there should be a more straightforward way to get to this.

For (d) it seems to me that there should be some function to satisfy these restrictions (since for $x < 0$ we can certainly let $f$ take on arbitrarily large negative values). It isn't clear to me how to systematically identify such a function though.

This problem comes from the exercises immediately following the statement and proofs of the first and second fundamental theorems of calculus, and a brief section on deducing properties of a function from its derivative; such as, a nonnegative derivative on an interval $\implies$ the function is increasing on the interval.

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    @ChristianBlatter Thanks for the suggestion. That was the basic idea I needed to arrive at the answer below. I had to play around with it a bit to get it to fit the given constraints.2012-03-21

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For part (c) I think the proof looks fine now.

For part(d): Prompted by one of the comments, I propose the following piecewise definition that I believe meets the requirements of the question: $f(x) = \begin{cases} x^2 + x + 1 & \text{if } x \geq 0,\\ -1/(x-1) & \text{if } x \leq 0. \end{cases}$

The graph of the function is given by:

enter image description here

Then we have $f''(x)$ exists and $>0$ for every $x$. Further, the other required conditions are easily verified.

If anyone has a better solution, or a systematic way to approach finding such a function, I'd still like to see such an answer. This was arrived at in a very ad hoc way (in that once I decided the vaguest properties each piece needed to have I just repeatedly adjusted the definition until it simultaneously met all of the requirements).