Every function that is continuous on a closed interval must have a maximum and a minimum value on the interval. These values may all be the same (y = 2 on [-3,3]); or the function may reach these values more than once (y = sin(x)).
If the function is defined on a closed interval, then the extreme values are either (1) at an endpoint of the interval or (2) at a critical number. This is known as the Extreme Value Theorem. Thus, one way of finding the extreme values is to simply find the value of the function at the endpoints and the critical points and compare these to find the largest and smallest. This is called the Candidates’ Test or the Closed Interval Test. It is a good one to “play” with: do some sketches of the different situation above; discuss why the interval must be closed.
On an open or closed interval, the shapes can change is if the first derivative is zero or undefined at the point where two shapes join. In this case the point is a local extreme value of the function – a local maximum or minimum value. Specifically:
- If the first derivative changes from positive to negative, the shape of the function changes from increasing to decreasing and the point is a local maximum. If the first derivative changes from negative to positive, the shape of the function changes from decreasing to increasing and the point is a local minimum.
This is a theorem called the First Derivative Test. By finding where the first derivative changes sign and in which direction it change (positive to negative, or negative to positive) we can locate and identify the local extreme value precisely.
- Another way to determine if a critical number is the location of a local maximum or minimum is a theorem called the Second Derivative Test.
If the first derivative is zero (and specifically not if it is undefined) and the second derivative is positive, then the graph has a horizontal tangent line and is concave up. Therefore, this is the location of a local minimum of the function.
Likewise, if the first derivative is zero at a point and the second derivative is negative there, the function has a local maximum there.
If both the first and second derivatives are zero at a point, then the second derivative test cannot be used, for example y = x4 at the origin.
The mistake student make with the second derivative test is in not checking that the first derivative is zero. If “justify your answer” is required, students should be sure to show that the first derivative is zero as well as the sign of the second derivative.
In the case where both the first and second derivatives are zero at the same point the function changes direction but not concavity (e.g. f (x) = x4 at the origin), or changes concavity but not direction (e.g. f (x) = x3 at the origin).
This is a revised version of a post published on October 22, 2012