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 [latex]f[/latex] that is twice-differentiable on an open interval [latex]I[/latex] containing [latex]a[/latex].
If [latex]f^{\prime \prime}(x)<0[/latex] and [latex]f^{\prime}(a)=0[/latex], [latex]f[/latex]is concave down at [latex]a[/latex], indicating a local maximum.
If [latex]f^{\prime \prime}(x)>0[/latex] and [latex]f^{\prime}(a)=0[/latex], [latex]f[/latex]is concave up at [latex]a[/latex], suggesting a local minimum at [latex]a[/latex].
Furthermore, if [latex]f^{\prime \prime}[/latex] is continuous over [latex]I[/latex] and remains positive, [latex]f[/latex] is consistently concave up across [latex]I[/latex], which helps in determining the behavior of [latex]f[/latex] at other critical points.
For instance, suppose there exists a point [latex]b[/latex] such that [latex]f^{\prime}(b)=0[/latex] and [latex]f^{\prime \prime}[/latex] is positive throughout, [latex]f[/latex] has a local minimum at [latex]b[/latex]. 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 [latex]f[/latex] such that [latex]f^{\prime \prime}[/latex] is continuous. Since [latex]f^{\prime}(a)=0[/latex] and [latex]f^{\prime \prime}(a)<0[/latex], there is an interval [latex]I[/latex] containing [latex]a[/latex] such that for all [latex]x[/latex] in [latex]I[/latex], [latex]f[/latex] is increasing if [latex]x<a[/latex] and [latex]f[/latex] is decreasing if [latex]x>a[/latex]. As a result, [latex]f[/latex] has a local maximum at [latex]x=a[/latex]. Since [latex]f^{\prime}(b)=0[/latex] and [latex]f^{\prime \prime}(b)>0[/latex], there is an interval [latex]I[/latex] containing [latex]b[/latex] such that for all [latex]x[/latex] in [latex]I[/latex], [latex]f[/latex] is decreasing if [latex]x<b[/latex] and [latex]f[/latex] is increasing if [latex]x>b[/latex]. As a result, [latex]f[/latex] has a local minimum at [latex]x=b[/latex].
second derivative test
Suppose [latex]f^{\prime}(c)=0, \, f^{\prime \prime}[/latex] is continuous over an interval containing [latex]c[/latex].
If [latex]f^{\prime \prime}(c)>0[/latex], then [latex]f[/latex] has a local minimum at [latex]c[/latex].
If [latex]f^{\prime \prime}(c)<0[/latex], then [latex]f[/latex] has a local maximum at [latex]c[/latex].
If [latex]f^{\prime \prime}(c)=0[/latex], then the test is inconclusive.
Note that for case iii. when [latex]f^{\prime \prime}(c)=0[/latex], then [latex]f[/latex] may have a local maximum, local minimum, or neither at [latex]c[/latex].
The functions [latex]f(x)=x^3[/latex], [latex]f(x)=x^4[/latex], and [latex]f(x)=−x^4[/latex] all have critical points at [latex]x=0[/latex]. In each case, the second derivative is zero at [latex]x=0[/latex].
However, the function [latex]f(x)=x^4[/latex] has a local minimum at [latex]x=0[/latex] whereas the function [latex]f(x)=−x^4[/latex] has a local maximum at [latex]x=0[/latex] and the function [latex]f(x)=x^3[/latex] does not have a local extremum at [latex]x=0[/latex].
Let’s now look at how to use the second derivative test to determine whether [latex]f[/latex] has a local maximum or local minimum at a critical point [latex]c[/latex] where [latex]f^{\prime}(c)=0[/latex].
Use the second derivative to find the location of all local extrema for [latex]f(x)=x^5-5x^3[/latex].
To apply the second derivative test, we first need to find critical points [latex]c[/latex] where [latex]f^{\prime}(c)=0[/latex].
The derivative is [latex]f^{\prime}(x)=5x^4-15x^2[/latex]. Therefore, [latex]f^{\prime}(x)=5x^4-15x^2=5x^2(x^2-3)=0[/latex] when [latex]x=0,\pm \sqrt{3}[/latex].
To determine whether [latex]f[/latex] has a local extrema at any of these points, we need to evaluate the sign of [latex]f^{\prime \prime}[/latex] at these points. The second derivative is
In the following table, we evaluate the second derivative at each of the critical points and use the second derivative test to determine whether [latex]f[/latex] has a local maximum or local minimum at any of these points.
[latex]x[/latex]
[latex]f^{\prime \prime}(x)[/latex]
Conclusion
[latex]−\sqrt{3}[/latex]
[latex]-30\sqrt{3}[/latex]
Local maximum
[latex]0[/latex]
[latex]0[/latex]
Second derivative test is inconclusive
[latex]\sqrt{3}[/latex]
[latex]30\sqrt{3}[/latex]
Local minimum
By the second derivative test, we conclude that [latex]f[/latex] has a local maximum at [latex]x=−\sqrt{3}[/latex] and [latex]f[/latex] has a local minimum at [latex]x=\sqrt{3}[/latex]. The second derivative test is inconclusive at [latex]x=0[/latex].
To determine whether [latex]f[/latex] has a local extrema at [latex]x=0[/latex], we apply the first derivative test.
To evaluate the sign of [latex]f^{\prime}(x)=5x^2(x^2-3)[/latex] for [latex]x \in (−\sqrt{3},0)[/latex] and [latex]x \in (0,\sqrt{3})[/latex], let [latex]x=-1[/latex] and [latex]x=1[/latex] be the two test points. Since [latex]f^{\prime}(-1)<0[/latex] and [latex]f^{\prime}(1)<0[/latex], we conclude that [latex]f[/latex] is decreasing on both intervals and, therefore, [latex]f[/latex] does not have a local extrema at [latex]x=0[/latex] as shown in the following graph.
Figure 10. The function [latex]f[/latex] has a local maximum at [latex]x=−\sqrt{3}[/latex] and a local minimum at [latex]x=\sqrt{3}[/latex]
Watch the following video to see the worked solution to this example.
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