Find a function with continuous second derivative satisfying given conditions

In each of the following cases find a function $f$ with continuous second derivative $f''$ satisfying the given conditions.

$f''(x) > 0$ for all $x$ , $f'(0) = 1$ , and $f'(1) = 0$ .
$f''(x) > 0$ for all $x$ , $f'(0) = 1$ , and $f'(1) = 3$ .
$f''(x) > 0$ for all $x$ , $f'(0) = 1$ , and $f(x) \leq 100$ for all $x > 0$ .
$f''(x) > 0$ for all $x$ , $f'(0) = 1$ , and $f(x) \leq 100$ for all $x < 0$ .

There can be no function meeting all of these conditions since $f''(x) > 0$ implies $f'(x)$ is an increasing function (since its derivative, $f''$ , is positive). But then $f'(0) > f'(1)$ contradicts that $f'(x)$ is increasing.
Let $f(x) = x^2 + x$ . Then
$\begin{align*} f'(x) = 2x + 1 && \implies f'(0) &= 1, \\ && \implies f'(1) &= 3 \end{align*}$

Furthermore, $f''(x) = 2 > 0$ for all $x$ .
There can be no function meeting all of these conditions. Again, $f''(x) > 0$ for all $x$ implies that $f'(x)$ is increasing for all $x$ . Therefore, $f'(0) = 1$ implies $f'(x) > 1$ for all $x > 0$ . Then, by the mean-value theorem, we know that for any $b > 0$ there exists some $c \in (0,b)$ such that
$f(b) - f(0) = f'(c) (b-0) \quad \implies \quad f(b) > b + f(0) \qquad (\text{since } f'(c) > 1).$

Now, choose $b > 100 - f(0)$ . Then, $f(b) > 100$ , contradicting that $f(x) \leq 100$ for all $x$ .
We’ll define $f$ piecewise as follows
$f(x) = \begin{dcases} 1 + x + x^2 & \text{if } x \geq 0 \\ \frac{1}{1-x} & \text{if } x < 0. \end{dcases}$

Then, we can take the derivative of each piece (and see that they are equal, so the derivative is defined at $x = 0$ )

$f'(x) = \begin{dcases} 1 + 2x &\text{if } x \geq 0 \\ \frac{1}{(1-x)^2} & \text{if } x < 0. \end{dcases}$

Taking the derivative again we find

$f''(x) = \begin{dcases} 2 & \text{if } x \geq 0 \\ \frac{2}{(1-x)^3} & \text{if } x < 0. \end{dcases}$

Thus, $f''(x) > 0$ for all $x$ and $f'(0) = 1$ . Furthermore, for $x < 0$ we have

$f(x) = \frac{1}{1-x} \leq 100 \text{ for all } x < 0.$