What are the points of inflection, if any, of #f(x)=x^3 - 12x^2 #?

The first point is (0, 0). The second is far away. These are turning points.

#y''=6x-24=0, for x = 4. This point ( 4, -128) is a candidate for point of

inflexion (POI).

#y'''= 6. y''' not 0 when y'' = 0 is a sufficient condition,

for the POI to be this point (4, -128).

Employing the contracting scale 1 for 20 for y only and 1 for 2 in x,

this POI and additional characteristics not shown by this graph,

could be revealed from concealment.

This edition, I was able to achieve contraction in a different graph (the

) to expose additional features. However, for the POI (point of

graph{y=x^3-12x^2 [-9.915, 10.085, -9.52, 0.48]} and graph{y=x^3-12x^2 [-324.5, 324.5, -166.8, 157.7]}

To find the points of inflection, we first find the second derivative of the function. The second derivative helps us determine the concavity of the function, which is necessary to identify inflection points.

First derivative: f'(x) = 3x^2 - 24x Second derivative: f''(x) = 6x - 24

To find the points of inflection, we set the second derivative equal to zero and solve for x:

6x - 24 = 0 6x = 24 x = 4

Now, we test the concavity of the function around x = 4:

For x < 4, choose x = 3: f''(3) = 6(3) - 24 = 18 - 24 = -6 (negative, so concave down)

For x > 4, choose x = 5: f''(5) = 6(5) - 24 = 30 - 24 = 6 (positive, so concave up)

Since the concavity changes at x = 4, the point (4, f(4)) is an inflection point.