Define ismax(f, x, c) = ((d(f,x)x=c) = 0) and ((d(f, x, 2)x=c) < 0)Define ismin(f, x, c) = ((d(f,x)x=c) = 0) and ((d(f, x, 2)x=c) > 0)
(It is easy to get the second function by simply editing the first one. If you are "shaky" on entering userdefined functions into your calculator, click here.) These functions are boolean functions  they return "true" or "false" as their values. You may also find it useful to have a function that just calculates the value of the second derivative at a particular value of x:
Define d2at(f, x, c) = d(f, x, 2)x=c
This can save you a little typing, but you have to interpret the results for yourself.
Here is a solution for Example 4 on page 183 of the Larsen text "Find the relative extrema for f(x) = 3x^{5} + 5x^{3}." The first step is to find critical values of f. The screen shot at the right shows the result of using the critnum() function to find critical values at x = 1, x = 0, and x = 1. 

Using the ismax() function, you can see that x=1 produces a relative maximum, but x = 0 does not. 

Using the ismin() function shows that x = 0 is also not the location of a relative minimum (so it must produce a point of inflection). However, x = 1 does produce a relative minimum of f. 

Instead, you could have used the d2at() function. The screen shot at right shows the results for this example. Since the second derivative is negative at x = 1, it produces a relative maximum. The second derivative is zero at x = 0, so x = 0 is a point of inflection. Also, since the second derivative is positive at x = 1, it produces a relative minimum. 
