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Lecture 18: Concavity and The 2nd Derivative Test

Goal: Use the 2nd derivative to determine properties of the original function.

Concave Up [CU] vs Concave Down [CD]

Figure 1 shows an example of a graph that is concave up. Formally, a function is concave up if for any two points on the graph, the line segment connecting them is above the graph. However, informally, a function is concave up if it is bowed upward to resemble a cup or a bowl. If we draw a bunch of tangent lines, we see their slope are increasing as we read the graph from left to right. We saw last lecture is a function is increasing, its derivative is positive. Hence, if a function \(f\) is concave up, then \([f']' = f''>0\).

Curve concave up with chord above it and increasing tangent slopes at marked points — Figure 1

A hand-drawn sketch labeled Concave Up [CU]. Vertical and horizontal axes are drawn in black. A black curve dips down near the left and rises steeply to the right, forming a valley shape. Two blue dots are marked on the curve, one near the low left portion and one on the rising right portion, connected by a straight blue chord line that lies above the curve between them. Short red tangent line segments are drawn at several points along the curve, with small red circles marking the points of tangency; the tangent segments become steeper as one moves to the right, indicating that the slope is increasing.

Similarly, Figure 2 shows an example of a graph that is concave down. Formally, a function is concave down if for any two points on the graph, the line segment connecting them is below the graph. However, informally, a function is concave down if it is bowed downward to resemble a cap or an arch. Using similar reasoning, if \(f\) is concave down, then the slope of the tangents are decreasing. Hence, \([f']' = f'' < 0\).

Curve concave down with chord below it and decreasing tangent slopes at marked points — Figure 2

A hand-drawn sketch labeled Concave Down [CD]. Vertical and horizontal axes are drawn in black. A black curve rises steeply from the lower left and flattens as it moves right, forming an arch shape. Two blue dots are marked on the curve, one on the steep left portion and one near the flat top at the right, connected by a straight blue chord line that lies below the curve between them. Short red tangent line segments with small red circles mark points of tangency; the tangent segments become less steep moving to the right, indicating that the slope is decreasing.

We want a way to determine the concavity of a function on an interval. Ideally, without having to fiddle around with inequalities.

Theorem (C Test)

Let \( f \) be a function where \( f'' \) exists on an open interval \( I \)

If \( f''(x) > 0 \) for \( x \) in \( I \), \( f \) is CU on \( I \).
If \( f''(x) < 0 \) for \( x \) in \( I \), \( f \) is CD on \( I \).

There is a diagram to help remember the results of the C Test.

Smiley Face Diagram for remembering the Concavity Test — Figure

On the left, there is a smiley face with two plus sign as eyes. The smiley face on the left indicates "When the second derivative is positive" using the eyes and "the function is concave up" using the mouth. Likewise, there is a frownie face on the right with two negative signs as eyes. The frownie face on the right indicates "When the second derivative is negative" using the eyes and "the function is concave down" using the mouth.

Example 1

Discuss the shape of \( f(x) = x^3 - 12x + 1 \)

Step 1 Find when \( f' = 0 \) and \( f'' = 0 \).

\begin{align*} f'(x) =&~ 3x^2 - 12\\[0.5em] =&~ 3(x-2)(x+2) \overset{\text{set}}{=} 0 \\[0.5em] \Rightarrow x =&~ \pm 2 \end{align*} \begin{align*} f''(x) = 6x \overset{\text{set}}{=} 0\\[0.5em] \Rightarrow x =&~ \pm 0 \end{align*}

Step 2 Make sign chart for both \( f' \) and \( f'' \)

Sign chart for \( f' \) and \( f'' \)
	\( (-\infty,-2) \)	\( (-2,0) \)	\( (0,2) \)	\( (2,\infty) \)
Test Points	\(-100\)	\(-1\)	\(1\)	\(100\)
Sign of \( 3(x-2)(x+2) = f' \)	\(+\)	\(-\)	\(-\)	\(+\)
Sign of \( 6x = f'' \)	\(-\)	\(-\)	\(+\)	\(+\)
Results of the ID Test	Inc	Dec	Dec	Inc
Results of C Test	CD	CD	CU	CU

Conclusion:

\( f \) is increasing and CD on \( (-\infty, -2) \)
\( f \) is decreasing and CD on \( (-2, 0) \)
\( f \) is decreasing and CU on \( (0, 2) \)
\( f \) is increasing and CU on \( (2, \infty) \)

We actually have enough information about the graph to make a very crude sketch of the graph of \(f\), as we have in Figure 3.

Sketch of the cubic with point P marked near the origin and x-intercepts near -2 and 2 — Figure 3

A hand-drawn sketch of the cubic function on x and y axes. The curve comes up from the lower left with an arrow pointing down at the far left, rises to a local maximum to the left of the labeled tick at \( -2 \), then decreases through a filled dot labeled P located just above the origin on the y-axis, continues decreasing to a local minimum between the origin and the labeled tick at \( 2 \), then rises steeply upward with an arrow at the upper right. Tick marks on the horizontal axis are labeled \( -2 \) and \( 2 \). The point P marks where the concavity changes from concave down to concave up.

We saw last lecture that when a function changed from increasing to decreasing, interesting things can happen (mainly there is a relative extrema there). Where the concavity changes is also a point of interest.

Definition: If \( f \) changes concavity at a point \( P \), then \( P \) is called an inflection point (IP).

The function in Example 1 has one inflection point, labeled as \("P"\) in Figure 3. There, it is the location of where the derivative reaches a minimum and starts increasing again. This causes, well, an inflection in the graph. As with relative extrema, we want a way to locate inflection points. To do so, we have an analog of Fermat's Theorem.

Theorem:

If \( (c, f(c)) \) is an inflection point, then either \( f''(c) = 0 \) or undefined.

Why? \( c \) is the location of a relative extrema for \(f'\). Then just use Fermat's Theorem.

NOTE:

As with Fermat's Theorem, the reverse is not true. For example, take \(x^4\) and \(\sqrt{|x|}\); their second derivative at \(0\) is \(0\) and undefined respectively. However, they do not have an inflection point at \(0\) since concavity does not change at \(0\).

Graph of y equals x to the fourth, a flat-bottomed curve concave up everywhere — Figure 4

A blue hand-drawn graph on horizontal and vertical axes showing the curve \( x^4 \), labeled at the upper right. The curve is symmetric about the vertical axis, nearly flat where it touches the horizontal axis at the origin, and rises steeply on both sides. It is concave up everywhere, illustrating that the second derivative equaling zero at a point does not guarantee an inflection point.

Graph of y equals the square root of absolute value of x, with a cusp at the origin — Figure 5

A blue hand-drawn graph on horizontal and vertical axes showing the curve \( \sqrt{|x|} \), labeled at the upper right. The curve forms a sharp cusp at the origin where it meets the horizontal axis, rising steeply on both the left and right sides and flattening as it moves outward, so both branches are concave down. This illustrates a point where the second derivative is undefined but no change of concavity occurs.

Example 2

Determine the locations of all inflection points for \( f(x) = 3x^4 - 4x^3 - 6x^2 + 12x + 1 \)

Step 1 Find when \( f'' \) is 0 or undefined

\[ f'(x) = 12x^3 - 12x^2 + 12x + 12 \]\[ \begin{aligned} f''(x) &= 12\cdot 3x^2 - 12\cdot 2x + 12\cdot 1 \\ &= 12(3x^2 - 2x + 1) \\ &= 12(x-1)(3x+1) \overset{\text{set}}{=\!=} 0 \end{aligned} \]

Thus, \( x = -\frac{1}{3}, 1 \) are the only possible locations.

Step 2 Make sign chart for \( f'' \)

Sign chart for \( f'' \)
	\( \left(-\infty, -\frac{1}{3}\right) \)	\( \left(-\frac{1}{3}, 1\right) \)	\( (1,\infty) \)
Test Points	\( -100 \)	\( 0 \)	\( 100 \)
Sign of 12	\(+\)	\(+\)	\(+\)
Sign of \( x-1 \)	\(-\)	\(-\)	+
Sign of \( 3x+1 \)	\(-\)	\(+\)	\(+\)
Sign of \( f'' \)	\(+\)	\(-\)	\(+\)
Results of C Test	CU	CD	CU

Conclusion: \( f \) has inflection points at both \( x = -\frac{1}{3} \) and \( x = 1 \)

Example 3

Find all inflection points of

\[ f(x) = \frac{1}{20}x^5 - \frac{1}{6}x^4 + \frac{1}{6}x^3 - x + 42 \]

Step 1

\[ f'(x) = \frac{1}{4}x^4 - \frac{2}{3}x^3 + \frac{1}{2}x^2 - 1 \] \[ \begin{aligned} f''(x) &= x^3 - 2x^2 + x \\ &= x(x^2 - 2x + 1) \\ &= x(x-1)^2 \overset{\text{set}}{=\!=} 0\\ \Rightarrow x &= 0, 1 \end{aligned} \]

Step 2

Sign chart for \( f'' \)
	\( (-\infty, 0) \)	\( (0,1)\)	\( (1, \infty) \)
Test Point	\( -100 \)	\( \frac{1}{2} \)	\( 100 \)
Sign of \( x \)	−	+	+
Sign of \( (x-1)^2 \)	+	+	+
Sign of \( f'' \)	−	+	+
Results of C Test	CD	CU	CU

Conclusion: \( f \) has only one IP at \( (0, f(0)) = (0, 42) \)

Locating and Classifying Relative Extrema

Concavity can be used to detect relative extrema. Let \( c \) be a critical number of \( f \). If \(f''(c) > 0\), then \(f\) is concave down near \(c\) implying there is a relative minimum at \(c\) (See Figure 6).

Concave up curve with a minimum at c on a number line — Figure 6

A hand-drawn sketch of a curve that dips down to a low point and rises again, forming a concave up shape. A small dot marks the lowest point of the curve. Below the curve is a horizontal number line with a single tick mark labeled c, positioned directly beneath the dot. This illustrates that when \( f''(c) > 0 \), the function \( f \) is concave up near \( c \), so \( c \) gives a relative minimum.

Likewise, if \(f''(c) < 0\), then \(f\) is concave down near \(c\) implying there is a relative minimum at \(c\) (See Figure 7).

Concave down curve with a maximum at c on a number line — Figure 7

A hand-drawn sketch of a curve that rises to a high point and falls again, forming a concave down arch. A small dot marks the highest point of the curve. Below the curve is a horizontal number line with a single tick mark labeled c, positioned directly beneath the dot. This illustrates that when \( f''(c) < 0 \), the function \( f \) is concave down near \( c \), so \( c \) gives a relative maximum.

Theorem (The 2^nd Derivative Test)

Let \( f'' \) be continuous near \( c \) and \( f'(c) = 0 \).

If \( f''(c) > 0 \), then \( f \) has a rel. min. at \( c \)
If \( f''(c) < 0 \), then \( f \) has a rel. max at \( c \)
If \( f''(c) = 0 \), the test is inconclusive. [I.e., the test failed and you'll need to pick another test. This is not the same thing as saying \(f\) has neither a relative max nor min at \(c\)]

Why (3)?

Take for example, \(x^4,~-x^4\), and \(x^3\). In each case, their first and second derivatives at \(0\) are \(0\). However, \(x^4\) has a relative minimum at \(0\), \(-x^4\) has a relative maximum at \(0\), and \(x^3\) has neither at \(0\) (See Figures 8-10). Hence, if \(f''(c) = 0\), then we gain no information about relative extrema at that point.

Hand-drawn sketch of a flattened U-shaped curve resembling \(x^4\) crossing coordinate axes. — Figure 8

A hand-drawn diagram on a white background shows a blue curve shaped like a flattened U, representing the function \(x^4\), labeled in blue at the upper right corner of the image. The curve is wide and nearly flat near the center, then rises steeply on both the left and right sides, curving upward past the top of the visible drawing area. A blue vertical line and a blue horizontal line intersect near the center of the curve, representing the y-axis and x-axis. A small red mark resembling the number zero or the letter O appears in the upper left corner, possibly labeling the origin. Below the drawing, partially visible black handwritten text appears to begin a word or phrase, though it is cut off and not fully legible.

Hand-drawn sketch of a downward-opening curve resembling \(-x^4\) on coordinate axes. — Figure 9

A hand-drawn blue ink sketch on a white background showing a coordinate system with a vertical axis and a horizontal axis intersecting near the upper portion of the image. A smooth, rounded curve resembling an inverted dome or hill shape passes through the intersection area, rising near the center and sloping downward steeply on both the left and right sides, consistent with the shape of a negative quartic function. To the right of the curve, near the horizontal axis, the label \(-x^4\) is handwritten, indicating the curve represents the function \(y = -x^4\). The overall sketch appears informal and freehand, likely illustrating the general shape and symmetry of this function around the vertical axis, with the curve concave downward throughout.

Hand-drawn sketch of the curve \(x^3\) showing an S-shaped curve through the origin with horizontal and vertical reference lines. — Figure 10

A hand-drawn blue sketch on a white background depicting the graph of \(x^3\), labeled at the top right. The drawing shows a roughly vertical S-shaped curve that passes through a central crossing point, rising steeply upward to the right and descending steeply downward to the left, consistent with the shape of a cubic function. Overlaid on this curve are two additional straight blue lines: one horizontal line extending from left to right near the middle of the image, and one vertical line extending from top to bottom, both intersecting at roughly the same central point as the curve, likely representing reference axes. Below the drawing there is some faint, partially visible handwritten text that is illegible.

Example 4

Locate and classify all relative extrema for \( f(x) = x^3 - 3x^2 \)

Step 1 Solve \( f'(x) = 0 \)

\[ f'(x) = 3x^2 - 6x = 3x(x-2) \overset{\text{set}}{=\!=} 0 \]

So relative extrema could occur at \( x = 0, 2 \).

Step 2 Determine the sign of \( f''(0) \) and \( f''(2) \)

\[ \begin{gathered} f''(x) = 6x - 6 \\[0.5em] f''(0) = -6 < 0 \\[0.5em] f''(2) = 6 > 0 \end{gathered} \]

Conclusion: \( f \) has a relative maximum at \( x = 0 \) and a relative minimum at \( x = 2 \).

Example 5

Locate and classify all relative extrema for \( f(x) = e^x(x^2 - 5x - 13) \)

Find when \(f'=0\)
\[ \begin{aligned} f'(x) &= e^x(x^2 - 5x - 13) + e^x(2x - 5) \\ &= e^x(x^2 - 3x - 18) \\ &= e^x(x - 6)(x + 3) \overset{\text{set}}{=\!=} 0 \Rightarrow x = -3, 6 \end{aligned} \]
Compute the sign of \( f''(-3) \) and \( f''(6) \)
\[ \begin{aligned} f''(x) &= e^x(x^2 - 3x - 18) + e^x(2x - 3) \\ &= e^x(x^2 - x - 21) \end{aligned} \] \[ \begin{gathered} f''(-3) = e^{-3}(9 + 3 - 21) = -9e^{-3} < 0 \\[0.5em] f''(6) = e^6(36 - 6 - 21) > 0 \end{gathered} \]

Conclusion: \(f\) has a relative maximum at \( x = -3 \) and relative minimum at \( x = 6 \).