Reconstruction of functions

The reconstruction of functions deals with the establishment of functional equations. Some reconstruction tasks require differential calculus.

Example

We are looking for a second degree function that has an intersection with the y axis at $(0|-3)$ and a maximum point at $H(3|2)$.

1. Function and derivation

   A second degree function is a quadratic function. This looks like this:
   $f(x)=ax^2+bx+c$
   
   The derivative is also needed:
   $f'(x)=2ax+b$
   
   The goal is now to find the variables $a$, $b$ and $c$ with the given points.

2. Set up equations

   The other information is now used to build equations.
   
   The intersection with the y-axis $S_y(0|-3)$ is inserted into the function $f(x)=ax^2+bx+c$:
   $f(0)=-3$
   $a\cdot0^2+b\cdot0+c=-3$
   $c=-3$
   
   The same happens with the maximum point at $H(3|2)$
   $f(3)=2$
   $a\cdot3^2+b\cdot3+c=2$
   $9a+3b+c=2$
   
   The derivative is zero at maximum points.
   $f'(3)=0$
   $2a\cdot3+b=0$
   $6a+b=0$

3. Solve equations

   The equations can be solved with a system of linear equations.
   
   1. $c=-3$
   2. $9a+3b+c=2$
   3. $6a+b=0$
   First of all, the insertion method lends itself by inserting the I. equation into II.
   
   1. $9a+3b-3=2$
   2. $6a+b=0$
   There are now several possibilities, but the insertion process makes sense here. First change and then use.
   
   1. $9a+3b-3=2$
   2. $6a+b=0\quad|-6a$
      $b=-6a$
   II in I
   $9a-18a-3=2\quad|+3$
   $-9a=5\quad|:(-9)$
   $a=-\frac59$
   

4. Specify function equation

   The following variables are already known:
   $a=-\frac59$ and $c=-3$
   $b$ can be calculated with one of the equations:
   $b=-6a$ $=-6\cdot(-\frac59)$ $=\frac{10}3$
   
   The variables are used and we get the function we are looking for.
   $f(x)=ax^2+bx+c$
   $f(x)=-\frac59x^2+\frac{10}3x-3$