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.

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Remember

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.
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Method

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

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.
    $f(x)=ax^2+bx+c$
    $f(x)=-\frac59x^2+\frac{10}3x-3$