Application: Maximums and Minimums -- 2nd Derivative Test
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."
This is where the so-called Second Derivative Test comes in.
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.
The key is that if a critical point of
f is a maximum, then
- The function,
f, MUST go from increasing to decreasing. - Therefore, the derivative,
f', must go from positive to negative, or in other wordsf'is decreasing. - Thus the derivative of the derivative,
f'', must be negative.
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:
fgoes from increasing to decreasing (you can see this by just noticing the slope of the tangent line goes from positive to negative).f'goes from positive to negative.- Therefore the derivative of the derivative,
f'', must be negative.
derivative(f') into the input bar to calculate the derivative of f'. They key thing to see is f'' is negative when x=1.
Now try the same thing, but with A near x=3 where f has a minimum. Notice that as you move A left and right of x=3 everything is reversed. In particular:
fgoes from decreasing to increasing.f'goes from negative to positive- Therefore the derivative of the derivative,
f'', must be positive.
f has a critical point at the input x0.
- If
f''(x0)<0thenf(x0)is a maximum. - If
f''(x0)>0then f(x0)is a minimum. - If
f''(x0)=0then f(x0)could be either a maximum or a minimum or it could be a saddle point. In other words, the Second Derivative Test is inconclusive.
- all three have critical points at x=0.
- 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 saddle point.
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.
In the next section we'll get into another application of the Second Derivative Test.