The first derivative test provides a systematic approach to identify local extrema, but in some cases, using the second derivative can be more straightforward. A function must have a local extremum at a critical point, but not all critical points are extremas.
Consider a function f that is twice-differentiable on an open interval I containing a.
If f′′(x)<0 and f′(a)=0, fis concave down at a, indicating a local maximum.
If f′′(x)>0 and f′(a)=0, fis concave up at a, suggesting a local minimum at a.
Furthermore, if f′′ is continuous over I and remains positive, f is consistently concave up across I, which helps in determining the behavior of f at other critical points.
For instance, suppose there exists a point b such that f′(b)=0 and f′′ is positive throughout, f has a local minimum at b. The second derivative thus confirms the nature of local extrema by providing insight into the concavity of the function at critical points.
Figure 9. Consider a twice-differentiable function f such that f′′ is continuous. Since f′(a)=0 and f′′(a)<0, there is an interval I containing a such that for all x in I, f is increasing if x<a and f is decreasing if x>a. As a result, f has a local maximum at x=a. Since f′(b)=0 and f′′(b)>0, there is an interval I containing b such that for all x in I, f is decreasing if x<b and f is increasing if x>b. As a result, f has a local minimum at x=b.
second derivative test
Suppose f′(c)=0,f′′ is continuous over an interval containing c.
If f′′(c)>0, then f has a local minimum at c.
If f′′(c)<0, then f has a local maximum at c.
If f′′(c)=0, then the test is inconclusive.
Note that for case iii. when f′′(c)=0, then f may have a local maximum, local minimum, or neither at c.
The functions f(x)=x3, f(x)=x4, and f(x)=−x4 all have critical points at x=0. In each case, the second derivative is zero at x=0.
However, the function f(x)=x4 has a local minimum at x=0 whereas the function f(x)=−x4 has a local maximum at x=0 and the function f(x)=x3 does not have a local extremum at x=0.
Let’s now look at how to use the second derivative test to determine whether f has a local maximum or local minimum at a critical point c where f′(c)=0.
Use the second derivative to find the location of all local extrema for f(x)=x5−5x3.
To apply the second derivative test, we first need to find critical points c where f′(c)=0.
The derivative is f′(x)=5x4−15x2. Therefore, f′(x)=5x4−15x2=5x2(x2−3)=0 when x=0,±√3.
To determine whether f has a local extrema at any of these points, we need to evaluate the sign of f′′ at these points. The second derivative is
f′′(x)=20x3−30x=10x(2x2−3).
In the following table, we evaluate the second derivative at each of the critical points and use the second derivative test to determine whether f has a local maximum or local minimum at any of these points.
x
f′′(x)
Conclusion
−√3
−30√3
Local maximum
0
0
Second derivative test is inconclusive
√3
30√3
Local minimum
By the second derivative test, we conclude that f has a local maximum at x=−√3 and f has a local minimum at x=√3. The second derivative test is inconclusive at x=0.
To determine whether f has a local extrema at x=0, we apply the first derivative test.
To evaluate the sign of f′(x)=5x2(x2−3) for x∈(−√3,0) and x∈(0,√3), let x=−1 and x=1 be the two test points. Since f′(−1)<0 and f′(1)<0, we conclude that f is decreasing on both intervals and, therefore, f does not have a local extrema at x=0 as shown in the following graph.
Figure 10. The function f has a local maximum at x=−√3 and a local minimum at x=√3
Watch the following video to see the worked solution to this example.
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