Dual of a convex function $f:\mathbb{R} \to \mathbb{R}$: existence, solution to ODE 
Let $f(x)$ be a smooth strictly convex (i.e. $f''(x)>0
 \,\,\,\,\,\forall x)$ funtion of $x\in \mathbb{R}$. Define the dual
  function $F(p)$ as $$F(p)=\max_x [px-f(x)].$$ 
  
  
*
  
*Make a sketch to explain the definition of $F(p)$ and its existence 
  for all $p\in \mathbb{R}$.
  
*Show that in general $p=f'(x)$, that $F'(p)=x$, and that $F(p)$ is
  strictly convex. Combining these, show that the dual of $F(p)$ is
  again $f(x)$.
  
*Compute the duals of $x^2/2$ and $x^4/4$ and verify the last    statement.
  
*Show that the solution to the ODE $(f')^3 +f'=x$ with $f(0)=0$ can
  be    written as $$f(x) = \max_p \left[  px - \frac{p^2}{2} -
>     \frac{p^4}{4}    \right].$$ You may assume that $f(x)$ is strictly
  convex.
  

OK, either this is a poorly worded question, or I am missing something, or both. For #1, I am assuming they must mean that $F(p)$ takes values in $\mathbb{R} \cup \{\infty\}$. Otherwise, take $p=0$ and $$f(x) = \int_0^x (\arctan \xi + 5) d\xi.$$ Then $f''>0$ but $f'$ is never zero, so the function achieves no max (correct me if I'm wrong).
I agree that if $f'(x)=p$ has a solution, then $F(p)$ must be finite, because $px-f(x)$ is strictly concave, and it has a unique critical point (if it has one at all) at which we have a global max. 
For #2, I find the wording to be confusing (what does the statement "$p=f(x)$" mean?). I guess they're just saying $F'(f'(x)) = x$. Anyway, I don't know how to show this. I do see that $F$ is at least weakly convex, since it is the upper envelope of weakly convex functions (i.e. affine functions of the form ap+b).
 A: 
poorly worded 

Welcome to applied mathematics.
I'm a little puzzled by the interpretation of "smooth strictly convex" as $f''>0$. To me, $x\mapsto  x^4$ is smooth and strictly convex. It also appears below in the question. 
You are right, in general the maximum in the definition of $F$ may end up being infinite supremum. A typical example is $x\mapsto \sqrt{x^2+1}$. There are two ways to deal with this problem: 


*

*Allow convex functions to take on the value $+\infty$ (typical  in convex analysis) 

*Assume that $f$ has superlinear growth: $f(x)/x\to\infty $ as $x\to\infty$,


My guess is that taking the second way is the  intention here. 

Show that in general $p=f'(x)$, that $F'(p)=x$

They mean the value of $x$ for which the maximum is attained in the definition of $F(p)$. In this context it is convenient to juggle $x$ and $p$ between being dependent and independent variables. 
The part $p=f'(x)$ follows at once from the definition. Note that $f'$ is strictly increasing and continuous, hence a homeomorphism. We have $F(f'(x)) = xf'(x)-f(x)$, where differentiating both sides  yields
$$F'(f'(x)) f''(x) = f''(x)$$
Hence $$F'(p)=x\tag{1}$$ when $f''(x)>0$.  Which, depending on the assumptions, could be always. If $f$ is only smooth & strictly convex, then $f''>0$ on a dense set; the equality $F'(p)=x$ holds on the image of this dense set under $f'$, which is a homeomorphism. Both sides of (1) are increasing functions of $p$, and the right-hand side is continuous in $p$, being the inverse of $f'$. Hence $F'(p)=x$ holds everywhere. Incidentally, we see that $F'(p)$ is strictly increasing, thus $F$ is strictly convex.
Part 3 is straightforward on the basis of the above formulas for derivatives. In part 4 they want you to consider the dual of $f$, identifying it as something explicit. 
