NOTE: You may view the original PDFs of the notes written in class by clicking on the link for the 1:30 Section or 2:30 Section, respectively.

Lecture 17: The I/D and 1st Derivative Test

Goal: Use the 1st derivative to determine properties of the original function.

Recall what it means for a function \( f \) to increase or decrease on an interval.

We say \(f\) is increasing if whenever \( x_1 < x_2 \), \( f(x_1) < f(x_2) \) (See Figure 1 for an example).

Graph of an increasing function f with points at x1 and x2 showing f(x1) less than f(x2).
Figure 1
A hand-drawn coordinate sketch of an increasing function labeled f. The vertical axis is marked with \( f(x_1) \) low down and \( f(x_2) \) higher up. The horizontal axis is marked with \( x_1 \) to the left and \( x_2 \) to the right. The curve is S-shaped, starting at a low point above \( x_1 \) and rising to a higher point above \( x_2 \), with dots marking the two points on the curve. Red tick marks along the curve indicate the rising direction, showing that as \( x \) increases the function value increases.

We say \(f\) is decreasing if whenever \( x_1 < x_2 \), \( f(x_1) > f(x_2) \) (See Figure 2 for an example).

Graph of a decreasing function f with points at x1 and x2 showing f(x1) greater than f(x2).
Figure 2
A hand-drawn coordinate sketch of a decreasing function labeled f. The vertical axis is marked with \( f(x_1) \) high up and \( f(x_2) \) lower down. The horizontal axis is marked with \( x_1 \) to the left and \( x_2 \) to the right. The curve starts at a high point above \( x_1 \) and falls to a lower point above \( x_2 \), with dots marking the two points on the curve. Red tick marks along the curve indicate the falling direction, showing that as \( x \) increases the function value decreases.

If we want to determine when a function increases or decreases using the definition, we would have to fiddle around with inequalities. Ideally, we would rather do a calculation to determine this. When \(f\) is differentiable, we have a procedure to determine when it increases or decreases motivated by the following theorem:

Theorem (The I/D Test)

Let \( f \) be a differentiable function on an open interval \( I \):

  1. If \( f'(x) > 0 \) for every \( x \) in \( I \), then \( f \) is increasing on \( I \).
  2. If \( f'(x) < 0 \) for every \( x \) in \( I \), then \( f \) is dereasing. on \( I \).

Example 1

When is \( f(x) = \dfrac{x^2}{1+x^2} \) increasing and decreasing?

Step 1 Determine all critical numbers of \( f \)

\[ \begin{gathered} f'(x) = \frac{2x(1+x^2) - (2x)x^2}{(x^2+1)^2} = \frac{2x + 2x^3 - 2x^3}{(x^2+1)^2} \\[0.5em] = \frac{2x}{(x^2+1)^2} \overset{\text{set}}{=} 0 \\[0.5em] 2x = 0 \\[0.5em] x = 0 \end{gathered} \]

Step 2 Use the critical numbers to divide the \( x \)-axis into regions to determine the sign of \( f' \).

Sign chart for \( f' \) with critical number \( x = 0 \)
\( (-\infty, 0) \) \( (0, \infty) \)
Test Point: \( -1 \) \( 1 \)
Sign of \( 2x \): \( - \) \( + \)
Sign of \( (x^2+1)^2 \) \( + \) \( + \)
Sign of \( f' \) \( - \) \( + \)
Results of ID Test Dec Inc

Conclusion: \( f \) is increasing on \( (0, \infty) \), while \( f \) is decreasing on \( (-\infty, 0) \). Moreover, by the First Derivative Test (defined later in the notes), \(f\) has a relative minimum at \(x = 0\).

NOTE:

People often just use a number line to organize this information. Here is an example of how it would look like if we used it in the previous example.

Hand-drawn blue sketch of a number line with a tick mark in the middle and 0 underneath the tick mark. To the left of 0, there is a point on the numberline labeled -1, while on the right of the number line there is a point labeled 1.
Figure 3
Hand-drawn blue sketch of a number line with a tick mark in the middle and 0 underneath the tick mark. To the left of 0, there is a point on the numberline labeled -1, while on the right of the number line there is a point labeled 1.
The main benefit to this is there is less writing involved, but we will use a table for this lecture. Since it lets us organize our thoughts and determine the sign of the derivative one part at a time.

Example 2

Let \( f \) be a polynomial such that

\[ f'(x) = (x-2)(x+1)(x-4) \]

Determine when \( f \) is increasing and decreasing.

Step 1

\[ \begin{gathered} (x-2)(x+1)(x-4) \overset{\text{set}}{=} 0 \\[0.5em] \Rightarrow x = 2, -1, 4 \Rightarrow x = -1, 2, 4 \end{gathered} \]

Step 2 We make our table, which is sometimes called a "sign chart."

Sign chart for \( f' \) with critical numbers \( x = -1, 2, 4 \)
\( (-\infty, -1) \) \( (-1, 2) \) \( (2,4) \) \((4, \infty)\)
Test Point \( -10^6 \) \( 0 \) \( 3 \) \( 10^6 \)
Sign of \( x+1 \) \( - \) \( + \) \( + \) \( + \)
Sign of \( x-2 \) \( - \) \( - \) \( + \) \( + \)
Sign of \( x-4 \) \( - \) \( - \) \( - \) \( + \)
Sign of \( f' \) \( - \) \( + \) \( - \) \( + \)
Results of ID Test Dec Inc Dec Inc

Conclusion: \( f \) is increasing on \( (-1, 2) \cup (4, \infty) \) and \( f \) is decreasing on \( (-\infty, -1) \cup (2, 4) \). Moreover, by the First Derivative Test (defined later in the notes), \(f\) has relative minima at \(x=-1\) and \(x = 4\), while \(f\) has a relative maximum at \(x=2\).

Hand-drawn number line with marks at -1, 2, 4 and a cubic-like curve crossing the axis.
Figure 6
A freehand sketch of a horizontal axis drawn slightly tilted upward to the right, ending in an arrowhead on the right. Three tick marks are drawn on the axis and labeled, from left to right, -1, 2, and 4. A continuous curve is sketched over the axis: it comes down steeply from above on the far left, crosses the axis just left of the -1 tick, dips into a rounded minimum below the axis between -1 and the next crossing, then rises steeply back up crossing the axis a little to the right of -1. The curve continues rising to a local maximum above the axis near the 2 tick, then decreases to a shallow local minimum above the axis near the 4 tick, and finally rises steeply upward to the right ending in an arrowhead. The curve stays above the axis to the right of its second crossing.

Example 3

When is \( f(x) = x^3 \) increasing and decreasing?

Step 1

\[ f'(x) = 3x^2 \overset{\text{set}}{=} 0 \Longrightarrow x = 0 \]

Step 2

Sign chart for \( f' \) with critical number \( x = 0 \)
\( (-\infty, 0) \) \( (0, \infty) \)
Test Point \( -1 \) \( 1 \)
Sign of \( f' \) \( + \) \( + \)
Results Inc Inc

Conclusion: \( f \) is increasing on \( (-\infty, \infty) \) as opposed to \( (-\infty, 0) \cup (0, \infty) \). Moreover, by the First Derivative Test (defined later in the notes), \(f\) has no relative extrema.

Small sketch of a cubic curve with an inflection point, rising through the axes.
Figure 7
A small hand-drawn sketch in blue and red showing a pair of crossed axes with an S-shaped cubic curve passing through the origin. The curve rises from lower left, flattens momentarily at the origin where the axes cross, and then continues rising to the upper right, illustrating that \( f(x) = x^3 \) is increasing everywhere and has neither a relative maximum nor a relative minimum at \( x = 0 \).

NOTE:

As a general guideline, you will want to use open intervals (i.e., use parantheses as opposed to brackets). This is so as the I/D Test can not be applied if the endpoints are involved since \(f'\) is no longer strictly positive or negative on the interval. However, you may need to include the endpoint if the sign of \(f'\) does not change at the critical number (like what we did in Example 3). In Example 3, we determined \(f\) is increasing on \((-\infty, 0) \cup (0, \infty)\) and we can conclude \(f\) is increasing on \((-\infty, \infty)\) from the original definition of increasing, not from the I/D Test.

This is a bit of a rant, but long story short, you only need to break the interval at the critical number if it changes from increasing to decreasing (or vice versa).

Hunting for Relative Extrema

The I/D Test gives us a way to determine the locations of relative extrema since they are located where the derivative changes sign (see the Figures below):

Hand-drawn curve rising then falling, with a peak at point c showing \(f'>0\) on the left and \(f'<0\) on the right.
Figure 8
A hand-drawn sketch on a horizontal baseline depicts an arch-shaped curve resembling a smooth hill or parabola. The curve starts at the lower left, rises steadily to a rounded peak marked with a filled black dot at the top, then descends to the lower right. A dotted vertical line drops straight down from the peak to the baseline, meeting it at a point labeled c. On the left side of the curve, an arrow points upward and to the right, indicating the curve is increasing, and below it the label reads f prime greater than 0, written as \(f'>0\). On the right side of the curve, an arrow points downward and to the right, indicating the curve is decreasing, and below it the label reads f prime less than 0, written as \(f'<0\). The overall drawing illustrates the concept of a critical point c where the derivative of a function changes sign from positive to negative, indicating a local maximum at the peak of the curve.
Hand-drawn parabola with minimum at point c, showing \(f'<0\) on the left and \(f'>0\) on the right.
Figure 9
A hand-drawn sketch of an upward-opening parabola (U-shaped curve) with a dotted vertical line marking its axis of symmetry. At the bottom of the curve, where it touches a horizontal line, there is a small circled point labeled c, indicating the minimum of the function. To the left of this point, a downward curving arrow points toward the horizontal line with the label \(f' < 0\) beneath it, indicating the function is decreasing there. To the right, an upward curving arrow points away from the horizontal line with the label \(f' > 0\) beneath it, indicating the function is increasing there. This illustrates the typical behavior of a derivative changing sign from negative to positive at a local minimum point c.
Hand-drawn sketch of two curves illustrating increasing then decreasing slopes around critical points, for finding maxima and minima.
Figure 10
A handwritten diagram with a partially cropped heading reading 'for max's/mins'. Two separate graphs are shown side by side, each with a horizontal axis line and a vertical dashed line marking a critical point c. Left graph: an upward curved arrow labeled above the curve on the left side, and the curve rises from left to right, crossing the dashed vertical line, with another upward arrow to the right of the dashed line; below the axis is written \( f' > 0 \), then \( c \), then \( f' > 0 \), indicating the derivative is positive on both sides of the critical point c, suggesting an inflection or increasing behavior throughout. Right graph: a downward arrow on the left above the curve, the curve starts relatively flat, dips downward after crossing the dashed vertical line, with another downward arrow on the right; below the axis is written \( f' < 0 \), then \( c \), then \( f' < 0 \), indicating the derivative is negative on both sides of the critical point c. The overall sketch is drawn in black ink on a white background, apparently illustrating the first derivative test for identifying local maxima and minima by examining the sign of the derivative before and after critical points.

Theorem (1st Derivative Test)

Let \( c \) be a critical number of a differentiable function \( f \).

  1. If \( f' \) switches from positive to negative at \( c \), then \( f \) has a relative maximum at \( c \).
  2. If \( f' \) switches from negative to positive at \( c \), then \( f \) has a relative minimum at \( c \).
  3. If \( f' \) does not switch signs at \( c \), then \( f \) has neither a rel. max nor rel. min at \( c \).

Example 4

The critical numbers of \( g(x) = x + 2\sin x \) on \( (0, 2\pi) \) are \( \frac{2\pi}{3} \) and \( \frac{4\pi}{3} \). Determine the locations of any relative extrema on the interval \( (0, 2\pi) \)

Step 1 We already know the critical numbers, but we will need to find the derivative for Step 2.

\[ g'(x) = 1 + 2\cos x \]

Step 2

Unit circle sketch marking two angles with terminal rays in the second and third quadrants.
Figure 11
A small hand-drawn unit circle in red with horizontal and vertical axes through the center. Two radii are drawn into the second and third quadrants, each ending at a marked point on the circle. The upper-left point is labeled \( \frac{2\pi}{3} \) and the lower-left point is labeled \( \frac{4\pi}{3} \), indicating the two critical numbers. This is ultimately used to show the test points lie in their respective intervals.
Sign chart for \( g' \) on \( (0, 2\pi) \)
\( (0, \frac{2\pi}{3} )\) \( (\frac{2\pi}{3}, \frac{4\pi}{3}) \) \( (\frac{4\pi}{3}, 2\pi )\)
Test Point \( \frac{\pi}{2} \) \( \pi \) \( \frac{3\pi}{2} \)
Sign of \( g' \) \( g'\left(\frac{\pi}{2}\right) = 1 \quad + \) \( g'(\pi) = -1 \quad - \) \( g'\left(\frac{3\pi}{2}\right) = 1 \quad + \)
ID Test Inc Dec Inc

Conclusion: By the First Derivative Test, \(g\) has a relative maximum at \(x = \frac{2\pi}{3}\) and a relative minimum at \(x = \frac{4\pi}{3}\)

.

Example 5

Classify and determine all relative extrema of the function \( f(x) = \frac{1}{8} e^{2x} \left( 4x^3 - 6x^2 - 6x + 3 \right) \).

Step 1 Determine all critical numbers

\[ \begin{aligned} f'(x) &= \frac{1}{4} e^{2x} \left( 4x^3 - 6x^2 - 6x + 3 \right) + \frac{1}{8} e^{2x} \left( 12x^2 - 12x - 6 \right) \\ &= e^{2x} \left( x^3 - \frac{3}{2}x^2 - \frac{3}{2}x + \frac{3}{4} \right) + e^{2x} \left( \frac{3}{2}x^2 - \frac{3}{2}x - \frac{3}{4} \right) \\ &= e^{2x} \left( x^3 - 3x \right) \\ &= e^{2x} x \left( x^2 - 3 \right) \quad \overset{\text{Set}}{=} 0\\ \Longrightarrow x &=~ 0, \pm \sqrt{3} \end{aligned} \]

Step 2 Make the sign chart, using \(\sqrt{3} \approx 1.7\) to determine the test points.

Number line with critical numbers \( -\sqrt{3}\), \( 0 \), and \( \sqrt{3} \), with test points and signs
\( (-\infty, -\sqrt{3}) \) \( (-\sqrt{3}, 0) \) \( (0, \sqrt{3}) \) \((\sqrt{3}, 0)\)
Test Points \( -10^6 \) \( -1 \) \(1\) \( 100 \)
Sign of \( e^{2x} \) \( + \) \( + \) \( + \) \( + \)
Sign of \( x^2 - 3 \) \( + \) \( - \) \( - \) \( + \)
Sign of \( x \) \( - \) \( - \) \( + \) \( + \)
Sign of \( f' \) \( - \) \( + \) \( - \) \( + \)

Conclusion: By the First Derivative Test, \(f\) has relative minima at \(x = \pm\sqrt{3}\) and a relative maximum at \(x = 0\)