Find function f(x) Find function f(x), where:
$$f(3)=3$$
$$f'(3)=3$$
$$f'(4)=4$$
$$f''(3) = \nexists$$ 
How to find function like this in general? What steps should I do?
 A: Take a function $|x|$ which has no derivative at 0. Integrate it to get a function $x\cdot|x|$ (yes, $1\over2$ is missing, but who cares) which has no second derivative at 0. Shift it to get $(x-3)|x-3|$ which has no second derivative at 3. Add some $const\cdot x^2$ to get $f'(4)-f'(3)=1$. Then add some $const\cdot x$ to get whatever derivative you like at 3. Then add a constant to get the desired value at 3.
A: I would take a family of functions with some free parameters (in this case, you have three equations, so you should be going for three parametrs) and then slowly adjust the parameters.


*

*Start with $A|x-3|+B$, which is a function that does not have a derivative at $3$.

*This means that $\int A|x-3|+Bdx$ (which has derivatives everywhere) has no second derivative at $3$

*Calculating the integral gives $$f_{A,B,C} = \begin{cases}-A(\frac{x^2}{2}+3x )+Bx +C& x\leq 3\\A(\frac{x^2}2 - 3x)+Bx + C & x\geq 3\end{cases}$$ which is a function that has no second derivative at $3$.

*Now you simply calculate $f_{A,B,C}(3)$, $f'_{A,B,C}(3)$ and $f'_{A,B,C}(4)$ to determine what the values of $A,B,C$ should be.


The values you get should be $$3=f_{A,B,C}(3) = -\frac{27}{2}A + 3B + C\\ 3=f'_{A,B,C}(3) = B\\
4=f'_{A,B,C}(4) = A+B$$
Which is easy to solve if you do it in the right order (first solve for $B$, then $A$, then $C$)
