You’ve gotten comfortable using the First Derivative Test in Calculus to find relative extrema, so do you really need the Second Derivative Test? Read more to find out the answer.
When you’re introduced to the Second Derivative Test, it is typically applied to polynomials, like the example below. And as you can see, the problem can be solved just as easily with the First Derivative Test.
It is worth noting that if you’re not great with fractions, the First Derivative Test may be challenging in this example. In order to complete the sign chart, you must plug in a fraction into f’(x) between -1 and 0 along with 0 and 1. Therefore, the Second Derivative Test may be easier.
But, in this example, the Second Derivative Test is inconclusive at x=0. How frustrating! So what do you do? You must go back to do the First Derivative Test.
It begs the question from many students: “Do I need to know the Second Derivative Test?”
The answer is yes. But not for polynomials like the above example. You need the Second Derivative Test for differential equations.
For example, the question may read: Let dy/dx= 2x – y.
a) How do we know that (1, 2) is a critical point for y=f(x)?
b) Is (1, 2) a relative minimum, relative maximum, or neither? Justify your answer.
For part (b), to determine if a relative minimum or maximum occur, you can’t create a sign chart. Yes, we can plug in x-values above and below x= 1 but we can’t determine what y-values to plug in at those points.
This is how you would solve that problem.
Even though your teacher’s exam may specify in the directions to use the Second Derivative Test for polynomials, the AP Calculus exam won’t tell you which Test to use. It’s important to be able to identify and apply both derivative tests to find relative extrema (minimums and maximums).
Justification for First Derivative Test on the AP Calculus Exam
Most free response questions about relative extrema on the AP Calculus exam require a justification.
Appropriate justification for the First Derivative Test could be “f has a relative max at x=3 because f’(x) changes from positive to negative at x=3.” The reader is looking for a Calculus-based justification, which means a connection to the derivative. Therefore, a justification that “f has a relative max at x=3 because f changes from increasing to decreasing at x=3” is not appropriate as there is no Calculus justification is involved; this Algebraic justification is a correct statement but is not going to earn credit on the AP exam. Make sure to be clear in your justifications and do not use pronouns like “it.” For example, a response “f has a relative max at x=3 because it changes from positive to negative at x=3” is not going to earn points as the reader doesn’t know if “it” refers to f(x), f’(x), etc.
Justification for Second Derivative Test on the AP Calculus Exam
Appropriate justification for the Second Derivative Test requires two parts: first, showing the point is a critical value or has a horizontal tangent. In other words, showing that the slope is zero at that point. Second, there must be a mention to the second derivative, relating the concavity to a min/max. If the function f(x) is concave up at the critical point, f(x) has a relative minimum at that point. If the function is concave down at the critical point, there is a relative maximum. Looking at the first example above, I included both parts in saying “f has a relative max at x=1 because f’(1)=0 and f”(1)<0.” The second example uses dy/dx instead of f’(x) so you can see both notations.
For more details on justifications on the AP exam, read my post How to Pass the AP Calculus Exam.
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