**Problem : ***f* (*x*) = 2*x*^{3} -3*x*^{2} - 4. Use the second derivative test to classify the critical points.

*f'*(*x*) = 6*x*^{2} - 6*x*;

*f'*(*x*) = 0 at

*x* = 0 and

*x* = 1.

*f''*(*x*) = 12*x* - 6;

*f''*(0) = - 6, so there is a local max at

*x* = 0.

*f''*(1) = 6, so there is a local min at

*x* = 1.

**Problem : **
Describe the concavity of *f* (*x*) = 2*x*^{3} -3*x*^{2} - 4 and find any inflection points.

*f''*(*x*) = 12*x* - 6, so

*f''*(*x*) = 0 when

*x* = . The sign of

*f''*(*x*) and the
concavity of

*f* are depicted below:

*f* has an inflection point at

*x* = because the concavity of the graph
changes there.

**Problem : ***f* (*x*) = *sin*(*x*). Use the second derivative test to classify the critical points on the interval
[0, 2*Π*].

*f'*(*x*) = - *cos*(*x*);

*f'*(*x*) = 0 at

*x* = and

*x* = .

*f''*(*x*) = - *sin*(*x*);

*f''*() = - 1, so

*f* has a local maximum there.

*f''*() = 1, so

*f* has a local minimum there.

**Problem : **
Describe the concavity of *f* and find any inflection point for *f* (*x*) = *sin*(*x*) on the
interval [0, 2*Π*].

*f''*(*x*) = - *sin*(*x*), so

*f''*(*x*) = 0 at

*x* = 0,

*x* = *Π*, and

*x* = 2*Π*.
The sign of

*f''*(*x*) and the concavity of

*f* are depicted below:

On the interval

[0, 2*Π*],

*f* only has an inflection point at

*x* = *Π*. If we were to
extend the graph in both directions,

*x* = 0 and

*x* = 2*Π* would also be points of
inflection.