The First Derivative Test

Corollary 3 of the Mean Value Theorem showed that if the derivative of a function is positive over an interval [latex]I[/latex] then the function is increasing over [latex]I.[/latex] On the other hand, if the derivative of the function is negative over an interval [latex]I,[/latex] then the function is decreasing over [latex]I[/latex] as shown in the following figure.

A continuous function [latex]f[/latex] has a local maximum at point [latex]c[/latex] if and only if [latex]f[/latex] switches from increasing to decreasing at point [latex]c.[/latex] Similarly, [latex]f[/latex] has a local minimum at [latex]c[/latex] if and only if [latex]f[/latex] switches from decreasing to increasing at [latex]c.[/latex] If [latex]f[/latex] is a continuous function over an interval [latex]I[/latex] containing [latex]c[/latex] and differentiable over [latex]I,[/latex] except possibly at [latex]c,[/latex] the only way [latex]f[/latex] can switch from increasing to decreasing (or vice versa) at point [latex]c[/latex] is if [latex]{f}^{\prime }[/latex] changes sign as [latex]x[/latex] increases through [latex]c.[/latex] If [latex]f[/latex] is differentiable at [latex]c,[/latex] the only way that [latex]{f}^{\prime }.[/latex] can change sign as [latex]x[/latex] increases through [latex]c[/latex] is if [latex]{f}^{\prime }(c)=0.[/latex] Therefore, for a function [latex]f[/latex] that is continuous over an interval [latex]I[/latex] containing [latex]c[/latex] and differentiable over [latex]I,[/latex] except possibly at [latex]c,[/latex] the only way [latex]f[/latex] can switch from increasing to decreasing (or vice versa) is if [latex]{f}^{\prime } (c)=0[/latex] or [latex]{f}^{\prime }(c)[/latex] is undefined. Consequently, to locate local extrema for a function [latex]f,[/latex] we look for points [latex]c[/latex] in the domain of [latex]f[/latex] such that [latex]{f}^{\prime } (c)=0[/latex] or [latex]{f}^{\prime }(c)[/latex] is undefined. Recall that such points are called critical numbers of [latex]f.[/latex]

Note that [latex]f[/latex] need not have a local extrema at a critical number. The critical points are candidates for local extrema only. In (Figure) , we show that if a continuous function [latex]f[/latex] has a local extremum, it must occur at a critical number, but a function may not have a local extremum at a critical number. We show that if [latex]f[/latex] has a local extremum at a critical number, then the sign of [latex]{f}^{\prime }[/latex] switches as [latex]x[/latex] increases through that point.

Using (Figure) , we summarize the main results regarding local extrema.

• If a continuous function [latex]f[/latex] has a local extremum, it must occur at a critical number [latex]c.[/latex]
• The function has a local extremum at the critical number [latex]c[/latex] if and only if the derivative [latex]{f}^{\prime }[/latex] switches sign as [latex]x[/latex] increases through [latex]c.[/latex]
• Therefore, to test whether a function has a local extremum at a critical number [latex]c,[/latex] we must determine the sign of [latex]{f}^{\prime }(x)[/latex] to the left and right of [latex]c.[/latex]

This result is known as the first derivative test .

We can summarize the first derivative test as a strategy for locating local extrema.

Now let’s look at how to use this strategy to locate all local extrema for particular functions.

Using the First Derivative Test to Find Local Extrema

Use the first derivative test to find the location of all local extrema for [latex]f(x)={x}^{3}-3{x}^{2}-9x-1.[/latex] Use a graphing utility to confirm your results.

Solution

Step 1. The derivative is [latex]{f}^{\prime }(x)=3{x}^{2}-6x-9.[/latex] To find the critical numbers, we need to find where [latex]{f}^{\prime }(x)=0.[/latex] Factoring the polynomial, we conclude that the critical numbers must satisfy

[latex]3({x}^{2}-2x-3)=3(x-3)(x+1)=0.[/latex]

Therefore, the critical numbers are [latex]x=3,-1.[/latex] Now divide the interval [latex](-\infty ,\infty )[/latex] into the smaller intervals [latex](-\infty ,-1),(-1,3)\text{ and }(3,\infty ).[/latex]

Step 2. Since [latex]{f}^{\prime }[/latex] is a continuous function, to determine the sign of [latex]{f}^{\prime }(x)[/latex] over each subinterval, it suffices to choose a point over each of the intervals [latex](-\infty ,-1),(-1,3)\text{ and }(3,\infty )[/latex] and determine the sign of [latex]{f}^{\prime }[/latex] at each of these points. For example, let’s choose [latex]x=-2,x=0,\text{ and }x=4[/latex] as test points.

Interval	Test Point	Sign of [latex]{f}^{\prime }(x)=3(x-3)(x+1)[/latex] at Test Point	Conclusion
[latex](-\infty ,-1)[/latex]	[latex]x=-2[/latex]	[latex](\text{+})(-)(-)=+[/latex]	[latex]f[/latex] is increasing.
[latex](-1,3)[/latex]	[latex]x=0[/latex]	[latex](\text{+})(-)(\text{+})=-[/latex]	[latex]f[/latex] is decreasing.
[latex](3,\infty )[/latex]	[latex]x=4[/latex]	[latex](\text{+})(\text{+})(\text{+})=+[/latex]	[latex]f[/latex] is increasing.

Step 3. Since [latex]{f}^{\prime }[/latex] switches sign from positive to negative as [latex]x[/latex] increases through [latex]1,f[/latex] has a local maximum at [latex]x=-1.[/latex] Since [latex]{f}^{\prime }[/latex] switches sign from negative to positive as [latex]x[/latex] increases through [latex]3,f[/latex] has a local minimum at [latex]x=3.[/latex] These analytical results agree with the following graph.

Using the First Derivative Test

Use the first derivative test to find the location of all local extrema for [latex]f(x)=5{x}^{1\text{/}3}-{x}^{5\text{/}3}.[/latex] Use a graphing utility to confirm your results.

Solution

Step 1. The derivative is

[latex]{f}^{\prime }(x)=\frac{5}{3}{x}^{-2\text{/}3}-\frac{5}{3}{x}^{2\text{/}3}=\frac{5}{3{x}^{2\text{/}3}}-\frac{5{x}^{2\text{/}3}}{3}=\frac{5-5{x}^{4\text{/}3}}{3{x}^{2\text{/}3}}=\frac{5(1-{x}^{4\text{/}3})}{3{x}^{2\text{/}3}}.[/latex]

The derivative [latex]{f}^{\prime }(x)=0[/latex] when [latex]1-{x}^{4\text{/}3}=0.[/latex] Therefore, [latex]{f}^{\prime }(x)=0[/latex] at [latex]x=\pm 1.[/latex] The derivative [latex]{f}^{\prime }(x)[/latex] is undefined at [latex]x=0.[/latex] Therefore, we have three critical numbers: [latex]x=0,[/latex] [latex]x=1,[/latex] and [latex]x=-1.[/latex] Consequently, divide the interval [latex](-\infty ,\infty )[/latex] into the smaller intervals [latex](-\infty ,-1),(-1,0),(0,1),[/latex] and [latex](1,\infty ).[/latex]

Step 2: Since [latex]{f}^{\prime }[/latex] is continuous over each subinterval, it suffices to choose a test point [latex]x[/latex] in each of the intervals from step 1 and determine the sign of [latex]{f}^{\prime }[/latex] at each of these points. The points [latex]x=-2,x=-\frac{1}{2},x=\frac{1}{2},\text{ and }x=2[/latex] are test points for these intervals.

Interval	Test Point	Sign of [latex]{f}^{\prime }(x)=\frac{5(1-{x}^{4\text{/}3})}{3{x}^{2\text{/}3}}[/latex] at Test Point	Conclusion
[latex](-\infty ,-1)[/latex]	[latex]x=-2[/latex]	[latex]\frac{(\text{+})(-)}{+}=-[/latex]	[latex]f[/latex] is decreasing.
[latex](-1,0)[/latex]	[latex]x=-\frac{1}{2}[/latex]	[latex]\frac{(\text{+})(\text{+})}{+}=+[/latex]	[latex]f[/latex] is increasing.
[latex](0,1)[/latex]	[latex]x=\frac{1}{2}[/latex]	[latex]\frac{(\text{+})(\text{+})}{+}=+[/latex]	[latex]f[/latex] is increasing.
[latex](1,\infty )[/latex]	[latex]x=2[/latex]	[latex]\frac{(\text{+})(-)}{+}=-[/latex]	[latex]f[/latex] is decreasing.

Step 3: Since [latex]f[/latex] is decreasing over the interval [latex](-\infty ,-1)[/latex] and increasing over the interval [latex](-1,0),[/latex] [latex]f[/latex] has a local minimum at [latex]x=-1.[/latex] Since [latex]f[/latex] is increasing over the interval [latex](-1,0)[/latex] and the interval [latex](0,1),[/latex] [latex]f[/latex] does not have a local extremum at [latex]x=0.[/latex] Since [latex]f[/latex] is increasing over the interval [latex](0,1)[/latex] and decreasing over the interval [latex](1,\infty ),f[/latex] has a local maximum at [latex]x=1.[/latex] The analytical results agree with the following graph.

Concavity and Points of Inflection

We now know how to determine where a function is increasing or decreasing. However, there is another issue to consider regarding the shape of the graph of a function. If the graph curves, does it curve upward or curve downward? This notion is called the concavity of the function.

(Figure) (a) shows a function [latex]f[/latex] with a graph that curves upward. As [latex]x[/latex] increases, the slope of the tangent line increases. Thus, since the derivative increases as [latex]x[/latex] increases, [latex]{f}^{\prime }[/latex] is an increasing function. We say this function [latex]f[/latex] is concave up . (Figure) (b) shows a function [latex]f[/latex] that curves downward. As [latex]x[/latex] increases, the slope of the tangent line decreases. Since the derivative decreases as [latex]x[/latex] increases, [latex]{f}^{\prime }[/latex] is a decreasing function. We say this function [latex]f[/latex] is concave down .

In general, without having the graph of a function [latex]f,[/latex] how can we determine its concavity? By definition, a function [latex]f[/latex] is concave up if [latex]{f}^{\prime }[/latex] is increasing. From Corollary 3, we know that if [latex]{f}^{\prime }[/latex] is a differentiable function, then [latex]{f}^{\prime }[/latex] is increasing if its derivative [latex]f''(x) \symbol{"3E} 0.[/latex] Therefore, a function [latex]f[/latex] that is twice differentiable is concave up when [latex]f''(x) \symbol{"3E} 0.[/latex] Similarly, a function [latex]f[/latex] is concave down if [latex]{f}^{\prime }[/latex] is decreasing. We know that a differentiable function [latex]{f}^{\prime }[/latex] is decreasing if its derivative [latex]f''(x)<0.[/latex] Therefore, a twice-differentiable function [latex]f[/latex] is concave down when [latex]f''(x)<0.[/latex] Applying this logic is known as the concavity test.

We conclude that we can determine the concavity of a function [latex]f[/latex] by looking at the second derivative of [latex]f.[/latex] In addition, we observe that a function [latex]f[/latex] can switch concavity ( (Figure) ). However, a continuous function can switch concavity only at a point [latex]x[/latex] if [latex]f''(x)=0[/latex] or [latex]f''(x)[/latex] is undefined. Consequently, to determine the intervals where a function [latex]f[/latex] is concave up and concave down, we look for those values of [latex]x[/latex] where [latex]f''(x)=0[/latex] or [latex]f''(x)[/latex] is undefined. When we have determined these points, we divide the domain of [latex]f[/latex] into smaller intervals and determine the sign of [latex]f''(x)[/latex] over each of these smaller intervals. If [latex]f''(x)[/latex] changes sign as we pass through a point [latex]x,[/latex] then [latex]f[/latex] changes concavity. It is important to remember that a function [latex]f[/latex] may not change concavity at a point [latex]x[/latex] even if [latex]f''(x)=0[/latex] or [latex]f''(x)[/latex] is undefined. If, however, [latex]f[/latex] does change concavity at a point [latex]a[/latex] and [latex]f[/latex] is continuous at [latex]a,[/latex] we say the point [latex](a,f(a))[/latex] is an inflection point of [latex]f.[/latex]

The Second Derivative Test

The first derivative test provides an analytical tool for finding local extrema, but the second derivative can also be used to locate extreme values. Using the second derivative can sometimes be a simpler method than using the first derivative.

We know that if a continuous function has a local extrema, it must occur at a critical number. However, a function need not have a local extrema at a critical number. Here we examine how the second derivative test can be used to determine whether a function has a local extremum at a critical number. Let [latex]f[/latex] be a twice-differentiable function such that [latex]{f}^{\prime }(a)=0[/latex] and [latex]f''(x)[/latex] is continuous over an open interval [latex]I[/latex] containing [latex]a.[/latex] Suppose [latex]f''(a)<0.[/latex] Since [latex]f''(x)[/latex] is continuous over [latex]I,[/latex] [latex]f''(x)<0[/latex] for all [latex]x\in I[/latex]. Then, by Corollary 3, [latex]{f}^{\prime }[/latex] is a decreasing function over [latex]I.[/latex] Since [latex]{f}^{\prime }(a)=0,[/latex] we conclude that for all [latex]x\in I,{f}^{\prime }(x) \symbol{"3E} 0[/latex] if [latex]x

Note that for case iii. when [latex]f''(x)(c)=0,[/latex] then [latex]f[/latex] may have a local maximum, local minimum, or neither at [latex]c.[/latex] For example, the functions [latex]f(x)={x}^{3},[/latex] [latex]f(x)={x}^{4},[/latex] and [latex]f(x)=-{x}^{4}[/latex] all have critical numbers 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,[/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 number [latex]c[/latex] where [latex]{f}^{\prime }(c)=0.[/latex]

Using the Second Derivative Test

Use the second derivative to find the location of all local extrema for [latex]f(x)={x}^{5}-5{x}^{3}.[/latex]

Solution

To apply the second derivative test, we first need to find critical numbers [latex]c[/latex] where [latex]{f}^{\prime }(c)=0.[/latex] The derivative is [latex]{f}^{\prime }(x)=5{x}^{4}-15{x}^{2}.[/latex] Therefore, [latex]{f}^{\prime }(x)=5{x}^{4}-15{x}^{2}=5{x}^{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''(x)[/latex] at these points. The second derivative is

[latex]f''(x)=20{x}^{3}-30x=10x(2{x}^{2}-3).[/latex]

In the following table, we evaluate the second derivative at each of the critical numbers 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''(x)[/latex]	Conclusion
[latex]-\sqrt{3}[/latex]	[latex]-30\sqrt{3}[/latex]	Local maximum
0	0	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)=5{x}^{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.

We have now developed the tools we need to determine where a function is increasing and decreasing, as well as acquired an understanding of the basic shape of the graph. In the next section we discuss what happens to a function as [latex]x\to \pm \infty .[/latex] At that point, we have enough tools to provide accurate graphs of a large variety of functions.