The issue we're up against is that after we find the critical points of a function f, it's unclear if they correspond to maximums or minimums of a function (or perhaps neither -- more on this later). If we can see a graph of the function, then it's all pretty obvious: just look at the graph. But not all functions have graphs that are easy to visualize, and even so, it's a good idea to have some tools in our tool box besides "just look at it."[br][br]This is where the so-called [b]Second Derivative Test[/b] comes in. [br][br]The Second Derivative Test provides a pretty straightforward process for testing critical points of a function to see if they are maximums or minimums. It isn't fool proof, but it's better than nothing. [br][br]The key is that if a critical point of [code]f[/code] is a maximum, then [br][list][*]The function, [code]f[/code], MUST go from increasing to decreasing.[/*][*]Therefore, the derivative, [code]f'[/code], must go from positive to negative, or in other words [code]f'[/code] is decreasing. [/*][*][b]Thus the derivative of the derivative,[code] f''[/code], must be negative.[/b][br][/*][/list][br]Pause and read this line of reasoning again. [br][br]Now take a look at the applet below, and move A just left and right of x=1 where f has a maximum.

Take your time here. As you move A back and forth, stop and think about each of the three bullet points. Notice that as you move A just left and right of x=1:[br][br][list][*][code]f[/code] goes from increasing to decreasing (you can see this by just noticing the slope of the tangent line goes from positive to negative).[/*][*][code]f'[/code] goes from positive to negative.[/*][*][b]Therefore the derivative of the derivative, [code]f''[/code], must be negative.[/b][/*][/list]Go ahead and type [code]derivative(f')[/code] into the input bar to calculate the derivative of [code]f'[/code]. They key thing to see is f'' is negative when x=1. [br][br]Now try the same thing, but with A near x=3 where [code]f[/code] has a minimum. Notice that as you move A left and right of x=3 everything is reversed. In particular:[br][br][list][*][code]f[/code] goes from decreasing to increasing.[/*][*][code]f'[/code] goes from negative to positive[/*][*][b]Therefore the derivative of the derivative, [code]f''[/code], must be positive.[/b][/*][/list][br]Let's summarize our findings. This summary is the [b]Second Derivative Test[/b]:[br][br]Suppose a function [code]f [/code]has a critical point at the input [code]x0[/code].[br][br][list][*]If [code]f''(x0)[/code][code]<0[/code] then [code]f(x0)[/code][code][/code] is a maximum. [/*][*]If [code]f''(x0)>0[/code] then [code]f(x0)[/code] is a minimum. [br][/*][*]If [code]f''(x0)=0[/code] then [code]f(x0)[/code] could be either a maximum or a minimum or it could be a saddle point. In other words, the Second Derivative Test is inconclusive. [br][/*][/list][br]Before we move on to using our new toy, let's just talk quickly about the last bullet point which shows us that this is NOT a foolproof test. Take a look at the following three functions. Notice:[br][br][list][*]all three have critical points at x=0. [br][/*][*]g has a minimum at x=0[/*][*]h has a maximum at x=0[/*][*]f has neither a maximum nor a minimum at x=0, and instead has what's called a [b]saddle point[/b].[/*][/list]

Now type the code snippets derivative(f,2) and derivative(g,2) and derivative(h,2) into the input bar to calculate the second derivatives of all three functions. Notice that each second derivative is 0 at x=0. This is the exact reason why the Second Derivative Test is inconclusive when the second derivative is 0. [br][br]In the next section we'll get into another application of the Second Derivative Test.