Skills Review for The Divergence and Integral Tests

Learning Outcomes

Calculate the limit of a function as 𝑥 increases or decreases without bound
Recognize when to apply L’Hôpital’s rule
Explain how the sign of the first derivative affects the shape of a function’s graph
State the first derivative test for critical points

In the Divergence and Integral Tests section, we will explore some methods that can be used to determine whether an infinite series diverges or converges. Here we will review how to take limits at infinity, L’Hopital’s Rule, and how to determine where a function is decreasing and increasing.

Take Limits at Infinity

(see Module 5, Skills Review for Sequences.)

Infinite Limits at Infinity

(see Module 5, Skills Review for Sequences.)

Apply L’Hôpital’s Rule

(see Module 5, Skills Review for Sequences.)

The First Derivative Test

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.

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Recall that $c$ is a critical point of a function $f$ if $f^{\prime}(c)=0$ or $f^{\prime}(c)$ is undefined.

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

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:

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

Example: Using the First Derivative Test to Find Increasing And Decreasing Intervals

Use the first derivative test to find all increasing and decreasing intervals for $f(x)=x^3-3x^2-9x-1$ .

Show Solution

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} - 2 x - 3) = 3 (x - 3) (x + 1) = 0

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

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

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.

Try It

Use the first derivative test to determine the increasing and decreasing intervals 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

$f$ is decreasing on the intervals $(-\infty, -2) and (3,\infty)$ and increasing on the interval $(-2,3)$ .

Example: Using the First Derivative Test to Find Increasing And Decreasing Intervals

Use the first derivative test to find the increasing and decreasing intervals for $f(x)=5x^{\frac{1}{3}}-x^{\frac{5}{3}}$ . Use a graphing utility to confirm your results.

Show Solution

The derivative is

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

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

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 $(1,\infty )$ .

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.

Sequences and Series: Prerequisite Material