- If on the interior of , then is increasing on .
- If on the interior of , then is decreasing on .

- Find the critical points of . Note that according to the definition in the text, critical points of are points where either is zero, the derivative doesn't exist, or endpoints of if is defined on a finite interval .
- The critical points divide the domain of into subintervals on which the sign of is constant. Check the sign of at one interior point on each subinterval. If it is positive, is increasing on that subinterval. If it is negative, is decreasing on that subinterval.

The second derivative, also provides information about the shape of the curve in terms of what is called concavity. Concavity can also be defined in several ways. Geometrically, it can be said that the graph of is concave up near a point if the tangent line at lies below the graph of on some open interval containing and is concave up if the tangent line lies above the graph of on some open interval containing . Algebraically, concavity is most often defined by saying that is concave up on an interval if is increasing on and is concave down on if is decreasing on . Using the theorem above and remembering that is the derivative of gives the following result.

- If on , then is concave up on .
- If on , then is concave down on .

Remember also that the second derivative can be helpful in determining local maximums and local minimums. That is, if you find a critical value, where or is undefined, then substitute the critical value into the second derivative. If the second derivative is positive, then there is a relative minimum there and if the second derivative is negative, then there is a relative maximum there.

> f := x-> x^3-3*x+1; > plot(f(x),x=-3..3); > solve(D(f)(x)=0,x); > D(f)(-2); > D(f)(0); > D(f)(2);The plot helps to see how many critical values you have. The

> D[1,1](f)(-1); > f(-1); > D[1,1](f)(1); > f(1);As you can see, the value of the second derivative at is negative implying that is a relative maximum. The value of the second derivative at is positive which means that is a relative minimum.

- For the function
, find the intervals on which it is increasing and the intervals on which it is decreasing.
- For the function
, find the intervals on which it is concave upward and the intervals on which it is concave downward.
- For the function , find all critical values first. Then classify each point as a relative maximum or a relative minimum using the second derivative test. Remember to include both the and coordinate for each point.
- Find the absolute extrema points for the function on the interval .

2001-12-11