Reconstruction

Reconstruction (it is also called curve fitting) is about building a function that fulfils certain conditions.

Some properties of the function are already known, and from these properties we are supposed to 'reconstruct' the function itself.

Example 1: maximum point and inflection point

Wanted is the polynomial function of degree three with the maximum point $H(2|4)$ and the inflection point $W(3|2)$.

Setup:

The degree tells you how many letters you are looking for:

Degree three means here: you are looking for $a$, $b$, $c$ and $d$. For $x$ you put in numbers.

The maximum point needs $f^{\prime }$ and the inflection point needs $f^{\prime \prime }$, so you differentiate with the letters already now:

Equations:

From every property you make an equation:

Look at the first two lines. $H(2|4)$ is a point with $x=2$ and $y=4$, and $f(x)$ is the $y$. So from $H(2|4)$ comes the equation $f(2)=4$. But $H$ is also a maximum point, and there the slope $m$ is equal to zero. $f^{\prime }(x)$ is the $m$, so $f^{\prime }(2)=0$ holds as well. That is why $H(2|4)$ stands there twice, once for each equation.

Four letters, four equations.

Solving:

For $f(x)$ the letters $a$, $b$, $c$ and $d$ are still missing, so you solve the system of equations.

Look at the column with the $d$. Above in III and in I the same $+d$ stands there, and when you subtract it falls away. In this way a letter disappears in every step, until only $a$ is left. This procedure is called the addition method.

With $a=1$ every equation gives the next letter:

Result:

Check:

Scheme

The same again in general, for every exercise of this kind:

Remember: as many equations as letters. Maximum point, minimum point and inflection point each give two equations, the saddle point gives three.

Every property becomes an equation

All the kinds that appear in the exercises, on one curve:

one equation

two equations

three equations

Maximum point and minimum point give the same two equations. Which of the two it is already stands in the exercise. $f^{\prime \prime }(2)<0$ is an inequality and does not come into the system of equations.

Your turn

Example 2: point-symmetric about the origin

Wanted is the polynomial function of degree three, point-symmetric about the origin, with the minimum point $T(2|-16)$.

Two letters, two equations.

With two equations, putting in is shorter than subtracting:

That is the substitution method, the second way through a system of equations.

The twice mirrored point $(-3|9)$ lies on the graph again, because $f(-3)=-27+36=9$. In $f(x)=-f(-x)$ both are contained: the inner minus in $f(-x)$ mirrors at the y-axis, the outer minus in front of it at the x-axis.

Example 3: axis-symmetric about the y-axis

Wanted is the polynomial function of degree four, axis-symmetric about the y-axis, with the maximum point $H(2|25)$ and a zero at $x=3$.

$e$ stays, because $e=e\cdot x^0$ and $0$ is even.

Three letters, three equations.

Axis-symmetric means mirroring once, at the y-axis: $f(-x)=f(x)$.

Example 4: saddle point

Wanted is the polynomial function of degree three with the saddle point $S(2|3)$. It meets the y-axis at $-5$.

Four letters, four equations. $S(2|3)$ gives three of them.

Exercises

Solutions to Your turn

Theory: why the number of conditions fits

As soon as a number is put in for $x$, only the letters are left:

Every condition gives exactly one such equation, and every letter needs one.

Symmetry deletes letters, it gives no equation. If an equation is missing at the end, then usually the y-value of a point is missing.