The First Derivative Test

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

Figure 1. Both functions are increasing over the interval $(a,b)$. At each point $x$, the derivative $f^{\prime}(x)>0$. Both functions are decreasing over the interval $(a,b)$. At each point $x$, the derivative $f^{\prime}(x)<0$.

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

Note that $f$ need not have a local extrema at a critical point. The critical points are candidates for local extrema only. In Figure 2, we show that if a continuous function $f$ has a local extremum, it must occur at a critical point, but a function may not have a local extremum at a critical point. We show that if $f$ has a local extremum at a critical point, then the sign of $f^{\prime}$ switches as $x$ increases through that point.

Figure 2. The function $f$ has four critical points: $a,b,c$, and $d$. The function $f$ has local maxima at $a$ and $d$, and a local minimum at $b$. The function $f$ does not have a local extremum at $c$. The sign of $f^{\prime}$ changes at all local extrema.

Using Figure 2, we summarize the main results regarding local extrema.

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

This result is known as the first derivative test.

First Derivative Test

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Suppose that $f$ is a continuous function over an interval $I$ containing a critical point $c$. If $f$ is differentiable over $I$, except possibly at point $c$, then $f(c)$ satisfies one of the following descriptions:

1. If $f^{\prime}$ changes sign from positive when [latex]xc[/latex], then $f(c)$ is a local maximum of $f$.
2. If $f^{\prime}$ changes sign from negative when [latex]xc[/latex], then $f(c)$ is a local minimum of $f$.
3. If $f^{\prime}$ has the same sign for [latex]xc[/latex], then $f(c)$ is neither a local maximum nor a local minimum of $f$.

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

Problem-Solving Strategy: Using the First Derivative Test

Consider a function $f$ that is continuous over an interval $I$.

1. Find all critical points of $f$ and divide the interval $I$ into smaller intervals using the critical points as endpoints.
2. Analyze the sign of $f^{\prime}$ in each of the subintervals. If $f^{\prime}$ is continuous over a given subinterval (which is typically the case), then the sign of $f^{\prime}$ in that subinterval does not change and, therefore, can be determined by choosing an arbitrary test point $x$ in that subinterval and by evaluating the sign of $f^{\prime}$ at that test point. Use the sign analysis to determine whether $f$ is increasing or decreasing over that interval.
3. Use the first derivative test and the results of step 2 to determine whether $f$ has a local maximum, a local minimum, or neither at each of the critical points.

Recall from Chapter 4.3 that when talking about local extrema, the value of the extremum is the y value and the location of the extremum is the x value.

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

Example: Using the First Derivative Test to Find Local Extrema

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

Show Solution

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

$3(x^2-2x-3)=3(x-3)(x+1)=0$

Therefore, the critical points are $x=3,-1$. Now divide the interval $(−\infty ,\infty)$ into the smaller intervals $(−\infty ,-1), \, (-1,3)$, and $(3,\infty)$.

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

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

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

Figure 3. The function $f$ has a maximum at $x=-1$ and a minimum at $x=3$

Watch the following video to see the worked solution to Example: Using the First Derivative Test to Find Local Extrema.

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Try It

Use the first derivative test to locate all local extrema for $f(x)=−x^3+\frac{3}{2}x^2+18x$.

Hint

Find all critical points of $f$ and determine the signs of $f^{\prime}(x)$ over particular intervals determined by the critical points.

Show Solution

$f$ has a local minimum at -2 and a local maximum at 3.

Example: Using the First Derivative Test

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

Show Solution

Step 1. The derivative is

$f^{\prime}(x)=\frac{5}{3}x^{-2/3}-\frac{5}{3}x^{2/3}=\frac{5}{3x^{2/3}}-\frac{5x^{2/3}}{3}=\frac{5-5x^{4/3}}{3x^{2/3}}=\frac{5(1-x^{4/3})}{3x^{2/3}}$.

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

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

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

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

Figure 4. The function f has a local minimum at $x=-1$ and a local maximum at $x=1$.

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 5(a) shows a function $f$ with a graph that curves upward. As $x$ increases, the slope of the tangent line increases. Thus, since the derivative increases as $x$ increases, $f^{\prime}$ is an increasing function. We say this function $f$ is concave up. Figure 5(b) shows a function $f$ that curves downward. As $x$ increases, the slope of the tangent line decreases. Since the derivative decreases as $x$ increases, $f^{\prime}$ is a decreasing function. We say this function $f$ is concave down.

Figure 5. (a), (c) Since $f^{\prime}$ is increasing over the interval $(a,b)$, we say $f$ is concave up over $(a,b)$. (b), (d) Since $f^{\prime}$ is decreasing over the interval $(a,b)$, we say $f$ is concave down over $(a,b)$.

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

We conclude that we can determine the concavity of a function $f$ by looking at the second derivative of $f$. In addition, we observe that a function $f$ can switch concavity (Figure 6). However, a continuous function can switch concavity only at a point $x$ if $f^{\prime \prime}(x)=0$ or $f^{\prime \prime}(x)$ is undefined. Consequently, to determine the intervals where a function $f$ is concave up and concave down, we look for those values of $x$ where $f^{\prime \prime}(x)=0$ or $f^{\prime \prime}(x)$ is undefined. When we have determined these points, we divide the domain of $f$ into smaller intervals and determine the sign of $f^{\prime \prime}$ over each of these smaller intervals. If $f^{\prime \prime}$ changes sign as we pass through a point $x$, then $f$ changes concavity. It is important to remember that a function $f$ may not change concavity at a point $x$ even if $f^{\prime \prime}(x)=0$ or $f^{\prime \prime}(x)$ is undefined. If, however, $f$ does change concavity at a point $a$ and $f$ is continuous at $a$, we say the point $(a,f(a))$ is an inflection point of $f$.

Figure 6. Since $f^{\prime \prime}(x)>0$ for $x<a$, the function $f$ is concave up over the interval $(−\infty,a)$. Since $f^{\prime \prime}(x)<0$ for $x>a$, the function $f$ is concave down over the interval $(a,\infty)$. The point $(a,f(a))$ is an inflection point of $f$.

Example: Testing for Concavity

For the function $f(x)=x^3-6x^2+9x+30$, determine all intervals where $f$ is concave up and all intervals where $f$ is concave down. List all inflection points for $f$. Use a graphing utility to confirm your results.

Show Solution

To determine concavity, we need to find the second derivative $f^{\prime \prime}(x)$. The first derivative is $f^{\prime}(x)=3x^2-12x+9$, so the second derivative is $f^{\prime \prime}(x)=6x-12$. If the function changes concavity, it occurs either when $f^{\prime \prime}(x)=0$ or $f^{\prime \prime}(x)$ is undefined. Since $f^{\prime \prime}$ is defined for all real numbers $x$, we need only find where $f^{\prime \prime}(x)=0$. Solving the equation $6x-12=0$, we see that $x=2$ is the only place where $f$ could change concavity. We now test points over the intervals $(−\infty ,2)$ and $(2,\infty)$ to determine the concavity of $f$. The points $x=0$ and $x=3$ are test points for these intervals.

Interval	Test Point	Sign of $f^{\prime \prime}(x)=6x-12$ at Test Point	Conclusion
$(−\infty ,2)$	$x=0$	$-$	$f$ is concave down
$(2,\infty )$	$x=3$	$+$	$f$ is concave up.

We conclude that $f$ is concave down over the interval $(−\infty ,2)$ and concave up over the interval $(2,\infty)$. Since $f$ changes concavity at $x=2$, the point $(2,f(2))=(2,32)$ is an inflection point. Figure 7 confirms the analytical results.

Figure 7. The given function has a point of inflection at $(2,32)$ where the graph changes concavity.

Watch the following video to see the worked solution to Example: Testing for Concavity.

We now summarize, in the box below, the information that the first and second derivatives of a function $f$ provide about the graph of $f$, and illustrate this information in Figure 8.

Figure 8. Consider a twice-differentiable function $f$ over an open interval $I$. If $f^{\prime}(x)>0$ for all $x \in I$, the function is increasing over $I$. If $f^{\prime}(x)<0$ for all $x \in I$, the function is decreasing over $I$. If $f^{\prime \prime}(x)>0$ for all $x \in I$, the function is concave up. If $f^{\prime \prime}(x)<0$ for all $x \in I$, the function is concave down on $I$.