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 user-defined 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) = -3x5 + 5x3." 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. |
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Using the ismax() function, you can see that x=1 produces a relative maximum, but x = 0 does not. |
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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. |
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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. |
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