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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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Your proof for c) looks correct at first glance. For d), how about $e^x$? – Alex Becker Mar 19 '12 at 4:16
@AlexBecker Sorry, I don't know what is $e^x$ at this point... next chapter :) – user23784 Mar 19 '12 at 4:17
You aren't familiar with the exponential function, or you don't know what it's derivative is? – Alex Becker Mar 19 '12 at 4:20
We have not yet defined what is an exponential function or a logarithm. Which is to say, I know what $e^x$ is, but it is not at my disposal yet. We have power functions (for rational powers) and trigonometric functions. – user23784 Mar 19 '12 at 4:22
@rar: Concerning (d): If you don't know about $\exp$ yet you could try a hyperbola with asymptotes $y=0$ and $x=1$. – Christian Blatter Mar 19 '12 at 9:08
up vote 1 down vote accepted

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).

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