# What is the high point of mathematics

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In another example, we calculate what proportion of the total population fell ill after the onset of a cold wave over the course of 15 days.

Status: May 13, 2013 | archive

A function that was determined empirically is used for this purpose: p (t) = 0.005 (15tht2t3) for the interval 1 ≤ t ≤ 15. It shows what proportion of the total population fell ill after the onset of a cold wave over the course of 15 days.

What percentage of the total population fell ill after the onset of a cold wave over the course of 15 days?

A first impression of the percentage of sick people can be obtained from a table of values: Between the 8th and 13th day the number of sick people reaches its peak, on the 15th day it is practically zero again due to the use of medication and / or immune defense.

The graph of the "disease function"

With many other values, one finally obtains the graph of the function p (t) = 0.005 (15t2 - t3). In this case, the most important point in terms of medical care for the population is when the proportion of sick people is greatest, i.e. where the graph reaches its maximum. Before the maximum, the function must increase, i.e. the derivative is positive, after the maximum the function has to decrease, i.e. the derivative is negative.

Graphical determination of the high point

It is obvious that the derivative is then zero at the maximum, i.e. the tangent slope is also zero, the tangent is parallel to the x-axis. The parallel can be drawn fairly precisely on the drawing board and it shows that the maximum is reached around the 10th day.

Calculation of the point in time of the maximum disease rate

A more precise result is obtained if the function p (t) is differentiated, it results p ’(t) = 0.015 t (10 - t). For p ’(t) = 0, the quadratic equation gives two solutions: t = 0 (this solution is ruled out because t is outside the definition range) and t = 10 .

The high point E of the function f (t)

If one puts t = 10 in the function p (t) = 0.005 (15t2 - t3) one, it results p = 2.5 . The proportion of the sick population therefore reaches a maximum of 2.5%.
The point E. with the coordinates (10 / 2.5) is now entered in the graph, it is called High point the function.

Calculation of the function f '(x) and its zeros - please click on the magnifying glass.

Now, in another function, 3rd degree all Find points with a horizontal tangent. The function equation is f (x) = (1/3)x3 – 3x2 + 8x - 2 . If you differentiate this function, you get the derivative function f '(x) = x2 - 6x + 8. For f '(x) = 0 you get the quadratic equation x2 - 6x + 8 = 0. She has the two solutions x1 = 4 andx2 = 2 .

Calculation of the coordinates of the two extreme points

Put these values ​​in the original functional equation f (x) = (1/3) x3 - 3x2 + 8x - 2 one gets y1 = 3,3 and y2 = 4,7 . So you get two extreme points for this function E.1(4/3,3) and E.2(2/4,7) .

What is the graph of the function?

These extreme values ​​are now entered in a coordinate system and it should be the task of the audience to draw the graph of the function.