Math Application differential calculus Reconstruction of functions

Reconstruction of functions

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



In the reconstruction of functions you look for a special function that satisfies given properties (e.g. type, points, slope, ...).
To do this, you set up equations and solve them using systems of equations.


  1. Function and derivation
  2. Set up equations
  3. Solve equations
  4. Specify function equation


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:

    The derivative is also needed:

    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 $f(x)=ax^2+bx+c$:

    The same happens with the maximum point at $H(3|2)$

    The derivative is zero at maximum points.
  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. $6a+b=0\quad|-6a$
    II in I
  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.