# Simplexety

#### General Curve Sketching (Curve Tracing)

If we examine the course of the function

#### f(x) = 1/9*x^3 - 1/3*x^2 - 8/3*x + 26/9

in the interval:

### x # [-10.000; 10.000]

then the function can be described by the following qualities and characteristic points:

## a) Derivatives of the Function

All necessary derivatives of the function are formed first:

### > f'(x) = 3*1/9*x^2-2*1/3*x-8/3

### > f''(x) = 6*1/9*x-2*1/3

### > f'''(x) = 6/9

Note: At present, Simplexety cannot shorten, which, of course, is desirable for mathematics.

## b) Local Extrema (Minima and Maxima)

To get a suitable condition, look at the course of the function f(x) and of the 1st and 2nd derivative f '(x) and f ''(x)

- Local Maximum: f '(x) = 0 (necessary condition) and f ''(x) < 0 (sufficient condition)
- Local Minimum: f '(x) = 0 (necessary condition) and f ''(x) > 0 (sufficient condition)

Results from it:

f '(x) = 3*1/9*x^2-2*1/3*x-8/3

f '(x) = 0

3*1/9*x^2-2*1/3*x-8/3 = 0

> Zero points (roots): x1=-2.0000 | x2=4.0000

Now the values of the zeros of the first derivative are successively used in the second derivative.

f ''(x) = 6*1/9*x-2*1/3

f ''(x1) = f ''(-2.0000) = -2.0000 < 0 --> Local Maximum

f ''(x1) = f ''(4.0000) = 2.0000 > 0 --> Local Minimum

After putting these zero points into the original function - the local extrema max/min, the [H]ighest and [L]owest points are:

### > H: max(-2.0000;6.0000)

### > L: min(4.0000;-6.0000)

## c) Inflection Points

If we examine the course of f(x) and f ''(x), it becomes evident that in the inflection point the 2nd derivative function has a value equal zero.

- Inflection Point: f ''(x) = 0 (necessary condition) and f '''(x) <> 0 (sufficient condition)

Thus, set the 2nd derivative to zero and if the 3rd derivative is equal to zero, then there must be a Inflection point. Whether the Inflection point is a Saddle point, that will be determined later on the position of the tangent.

f ''(x) = 6*1/9*x-2*1/3

f ''(x) = 0

6*1/9*x-2*1/3 = 0

Solution: x1=1.0000

Now each of the values of the zeros of the 2nd derivative are put into the 3rd derivative.

f '''(x) = 6/9

f '''(x1) = f '''(1.0000) = 0.6667 <> 0 --> Inflection Point (or Saddle Point)

After putting this zero point into the origin function, the inflection point therefore is:

### > WP: (1.0000;0.0000)

## d) Equation of the tangent at the inflection point

For the equation of the tangent f(xwp) = m xwp + b at the inflection point (tangent) calc at first the slope m

- The following applies: m = f '(xwp)

f '(x) = 3*1/9*x^2-2*1/3*x-8/3

Substituting xwp = 1.0000 yields:

### > m = f '(1.0000) = -3.0000

Conversion to b = f(xwp) - (m * xwp) results in b:

### > b = 3.0000

Result: The inflectional tangent function equation:

### > f(xwp) = -3.0000* xwp + 3.0000

Note: If m = f '(xwp) = 0 so, this is referred to as the Inflection point of the saddle point.

## e) Curvature Behavior (Concavity)

The Inflection point is either a L-R or R-L transition. Therefore exist intervals on which the graph of the function having a left or right curvature. Since the turn of the 2nd derivative has the value zero, there exists an interval (I), where the 3rd derivative is positive or negative.

Therefore, the condition:

- L-R Transition: f ''(x) = 0 and f '''(x) < 0
- R-L Transition: f ''(x) = 0 and f '''(x) > 0

With the above results of the inflection point, the type of transition is:

### > f '''(1.0000) = 0.6667 > 0 --> R-L Transition

## f) Zero Points (Roots of the Function)

The zeros are the points of intersection with the x-axis if the condition: f(x) = 0 is valid:

f(x) = 1/9*x^3 - 1/3*x^2 - 8/3*x + 26/9

f(x) = 0

1/9*x^3 - 1/3*x^2 - 8/3*x + 26/9 = 0

Solution:

### > Zeros of the function: x1=-4.1962 | x2=1.0000 | x3=6.1962