Applications of differentiation - the graph of a function and its derivatives
      Finding and classifying critical (or stationary) points, examples
Finding and classifying critical (or stationary) points examples
Example:  Find the extreme point of the quartic polynomial  y = x4 - 4x3 + 6x2 - 4x  and sketch its graph.
Solution:  To rewrite given polynomial from general into translatable form, and this way determine its source form, we should always first calculate the coordinates of translation
then, plug  x0 and  y0  into
 y + y0 = a4(x + x0)4 + a3(x + x0)3 + a2(x + x0)2 + a1(x + x0) + a0
to get the source form of the given polynomial
 y - 1 = 1 (x + 1)4 - 4(x + 1)3 + 6(x + 1)2 - 4(x + 1) + 0,  so the source function   y = x4.
Therefore, we can write,      y = x4 - 4x3 + 6x2 - 4x    or    y + 1 = (x - 1)4   or    f (x) = (x - 1)4 - 1.
To find the extreme point, calculate the first derivative of the above expression and set it to zero,
 f ' (x) = 4(x - 1)3,  then   f ' (x) = 0   or   4(x - 1)3 = 0   what yields  x = 1
as possible abscissa of the extreme point. 
To check if x = 1 is the abscissa of the extreme point, plug it into successive higher order derivatives, starting with the second derivative. Thus,
 f '' (x) = 12(x - 1)2,  since    f '' (1) = 0  we must proceed with the higher order derivatives, so
         f ''' (x) = 24(x - 1)  and   f ''' (1) = 0   and finally   f IV (x) = 24 > 0 
confirms x = 1 is the abscissa of the minimum as first of higher order derivatives that do not vanishes at this point is of even order and positive.
By plugging x = 1 into given quartic we find the minimum Tmin(1, -1), as shows the figure below.
Example:  Find extreme points and points of inflection of the quintic  y = -x5 + 5x4 - 7x3 + x2 + 4x -1 and sketch its graph.
Solution:  To rewrite given polynomial from general into translatable form, and this way determine its source form, we first calculate the coordinates of translation
then, plug  x0 and  y0  into
 y + y0 = a5(x + x0)5 + a4(x + x0)4 + a3(x + x0)3 + a2(x + x0)2 + a1(x + x0) + a0
to get the source form of the given polynomial
 y + 1 = - 1(x + 1)5 + 5(x + 1)4 - 7(x + 1)3 + (x + 1)2 + 4(x + 1) - 1,
what after expanding and reducing gives   y = - x5 + 3x3,    the source quintic.
Therefore, we can write   y = -x5 + 5x4 - 7x3 + x2 + 4x -1
      or     y - 1 = - (x - 1)5 + 3(x - 1)3    or     f(x) = - (x - 1)5 + 3(x - 1)3 + 1.
To find the extreme points, calculate the first derivative of the above expression set it to zero and solve for x
 f ' (x) - 5(x - 1)4 + 9(x - 1)2 - (x - 1)2 [5(x - 1)2 - 9]
then we set         f ' (x) = 0,       - (x - 1)2 [5(x - 1)2 - 9] = 0  
what yields,  x1 = 1x2 = 1 - 3/5  and  x3 = 1 + 3/5  as potential abscissas of the extreme points.
To check if  x1 = 1 is the abscissa of the extreme point, plug it into successive higher order derivatives, starting with the second derivative. Thus,
 f '' (x) - 20(x - 1)3 + 18(x - 1) - 2(x - 1) [10(x - 1)2 - 9]   since    f '' (1) = 0
we must proceed with the higher order derivatives,
 f ''' (x) =   - 60(x - 1)2 + 18   so that    f ''' (1) = 18 is not 0  
what classifies root  x1 = 1 as the abscissa of the point of inflection, as first of higher order derivatives that do not vanishes at this point is of odd order.
While roots,  x2 = 1 - 3/5  and  x3 = 1 + 3/5  plugged into the second derivative respectively yield
f '' (1 - 3/5) > 0,    so  x2 = 1 - 3/5   is the abscissa of the minimum   
            and       f '' (1 + 3/5) < 0,    so  x3 = 1 + 3/5   is the abscissa of the maximum.   
To find points of inflection of the given quintic we set the second derivative to zero and solve for x
 f '' (x) - 20(x - 1)3 + 18(x - 1),     - 2(x - 1) [10(x - 1)2 - 9]  = 0
what yields, x1 = 1x2 = 1 - 3/10  and  x3 = 1 + 3/10 as possible abscissas of the points of inflection.
The root  x1 = 1 we've already classified as the point of inflection.
Then we plug roots,  x2 = 1 - 3/10  and  x3 = 1 + 3/10  into the third derivative respectively
 f ''' (x) =   - 60(x - 1)2 + 18,   so     f ''' (1 - 3/10) ¹ 0   and     f ''' ( 1 + 3/10) is not 0
what classifies roots,  x2  and  xas the abscissas of the points of inflection.
By plugging the abscissas x1, x2 and x3 into the given quintic we get the ordinates of the points of inflection.
The graph of the given quintic polynomial and its source function is shown in the figure below.
Calculus contents D
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