Lesson 23 (10/23/26)

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Lesson 23 (10/23/26)

Today:

Review of first Derivative test
What does second derivative tell us
Second Derivative Test

Office Hours: MWF 130pm - 230pm

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Review: First Derivative Test

$Cartesian coordinate system showing a blue continuous function curve demonstrating various critical points and behaviors. From left to right: the curve increases with positive slope (red tangent segment) to a smooth local maximum with a horizontal red tangent line ($f'(c) = 0$); decreases with negative slope (red tangent segment) to a smooth local minimum with a horizontal tangent line; rises to another smooth local maximum with a horizontal tangent line; falls to another smooth local minimum with a horizontal tangent line; sharply rises to a sharp peak (corner/cusp where $f'(c)$ does not exist); drops steeply to a sharp trough marked with a red dot where $f'(c)$ DNE; and then increases steeply with a red tangent line indicating positive slope.$

Visual Description: Cartesian coordinate system showing a blue continuous function curve demonstrating various critical points and behaviors. From left to right: the curve increases with positive slope (red tangent segment) to a smooth local maximum with a horizontal red tangent line ($f'(c) = 0$); decreases with negative slope (red tangent segment) to a smooth local minimum with a horizontal tangent line; rises to another smooth local maximum with a horizontal tangent line; falls to another smooth local minimum with a horizontal tangent line; sharply rises to a sharp peak (corner/cusp where $f'(c)$ does not exist); drops steeply to a sharp trough marked with a red dot where $f'(c)$ DNE; and then increases steeply with a red tangent line indicating positive slope.

Critical points: $ f'(c) = 0 $ or $ f'(c) \text{ DNE} $

increasing: $ f' > 0 $

decreasing: $ f' < 0 $

if $ f' $ changes sign at $ x = c $ then we have a Relative/Local min/max

Relative max: $ f' $ changes from $ > 0 $ to $ < 0 $

Relative min: $ f' $ changes from $ < 0 $ to $ > 0 $

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eg 1

$ f(x) = x e^{-x} $, Find Relative max/min using 1st Derivative Test.

\[ f'(x) = \frac{d}{dx} (x e^{-x}) = e^{-x} + x \cdot e^{-x}(-1) = e^{-x}(1-x) \]

$ x = 1 $ is the only critical point

First derivative test number line partitioned at the critical point x = 1. To the left of 1, the derivative is evaluated as f' = e^(-x)(1 - x) > 0, indicated with an upward arrow showing that the function is increasing. To the right of 1, the derivative is evaluated as f' = e^(-x)(1 - x) < 0, indicated with a downward arrow showing that the function is decreasing. The transition from positive to negative derivative indicates a local maximum.

Visual Description: First derivative test number line partitioned at the critical point x = 1. To the left of 1, the derivative is evaluated as f' = e^(-x)(1 - x) > 0, indicated with an upward arrow showing that the function is increasing. To the right of 1, the derivative is evaluated as f' = e^(-x)(1 - x) < 0, indicated with a downward arrow showing that the function is decreasing. The transition from positive to negative derivative indicates a local maximum.

Relative Max.

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What does $ f''(x) $ tell us?

\[ f(x) \rightsquigarrow f'(x) = \frac{d}{dx} f(x) \quad \begin{aligned} &> 0 \implies f(x) \uparrow \\ &< 0 \implies f(x) \downarrow \end{aligned} \]

\[ f''(x) = \frac{d}{dx} (f'(x)) \rightsquigarrow \text{① } f''(x) > 0 \quad \text{then} \quad f'(x) \uparrow \]

\[ f' > 0 \qquad\qquad\qquad\qquad\qquad\qquad f' < 0 \]

Hand-drawn illustrations of concave up functions. The first curve is increasing with positive tangent slopes in red that become progressively steeper, showing f' is increasing. The second curve is decreasing with negative tangent slopes in red that flatten out towards zero, showing f' is increasing from more negative to less negative. The third curve is a full U-shaped concave up bowl. The fourth curve depicts two upward-curving arches meeting at a sharp peak.

Visual Description: Hand-drawn illustrations of concave up functions. The first curve is increasing with positive tangent slopes in red that become progressively steeper, showing f' is increasing. The second curve is decreasing with negative tangent slopes in red that flatten out towards zero, showing f' is increasing from more negative to less negative. The third curve is a full U-shaped concave up bowl. The fourth curve depicts two upward-curving arches meeting at a sharp peak.

We say graph is concave up

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2. $ f'' < 0 \sim \frac{d}{dx}(f'(x)) < 0 \Rightarrow f'(x) \text{ is } \downarrow $

Four sketches illustrating concave down curves where the second derivative is negative. From left to right: (1) A decreasing curve that bends downwards, with a red tangent line through a highlighted red point demonstrating a negative slope. (2) An increasing curve bending downwards, showing two red tangent lines at two points along the curve demonstrating that the slope becomes less steep as x increases. (3) Two concave-down curved segments that meet at a sharp downward-pointing cusp. (4) A smooth, symmetric inverted U-shaped parabola opening downwards with a crest at the top.

Visual Description: Four sketches illustrating concave down curves where the second derivative is negative. From left to right: (1) A decreasing curve that bends downwards, with a red tangent line through a highlighted red point demonstrating a negative slope. (2) An increasing curve bending downwards, showing two red tangent lines at two points along the curve demonstrating that the slope becomes less steep as x increases. (3) Two concave-down curved segments that meet at a sharp downward-pointing cusp. (4) A smooth, symmetric inverted U-shaped parabola opening downwards with a crest at the top.

if $ f'' < 0 \sim $ we say $ f(x) $ is Concave down.

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Second Derivative Test:

Relative max at $ x = c $:

Downward-opening concave curve illustrating a relative maximum at the peak, flanked by minus signs on the left and right sides to indicate negative curvature and concavity downwards.

Visual Description: Downward-opening concave curve illustrating a relative maximum at the peak, flanked by minus signs on the left and right sides to indicate negative curvature and concavity downwards.

$ f'(c) = 0 $
$ f''(c) < 0 $

eg: $ -x^2 $

Relative min at $ x = c $:

Upward-opening convex curve illustrating a relative minimum at the trough, flanked by plus signs on the left and right sides to indicate positive curvature and concavity upwards.

Visual Description: Upward-opening convex curve illustrating a relative minimum at the trough, flanked by plus signs on the left and right sides to indicate positive curvature and concavity upwards.

$ f'(c) = 0 $
$ f''(c) > 0 $

eg: $ x^2 $

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e.g.:

Apply Second Derivative test to Determine Relative max/min.

\[ f'(x) = 12x^3 - 12x^2 - 24x = 12x [x^2 - x - 2] \] \[ = 12x (x - 2)(x + 1) = 0 \]

points to check: $ x = 0, \, x = 2, \, x = -1 $

\[ f''(x) = 36x^2 - 24x - 24 = 12 [3x^2 - 2x - 2] \]

\[ f''(0) = -24 < 0 \]

Hand-drawn diagram of a concave-down curve with two negative signs above it, indicating that a negative second derivative represents a downward concavity and therefore a local maximum.

Visual Description: Hand-drawn diagram of a concave-down curve with two negative signs above it, indicating that a negative second derivative represents a downward concavity and therefore a local maximum.

$ x = 0 $ is Local Max

\[ f''(2) = 12 [12 - 4 - 2] = 72 > 0 \]

Hand-drawn diagram of a concave-up curve (U-shaped) with two plus signs above the endpoints, indicating that a positive second derivative represents an upward concavity and therefore a local minimum.

Visual Description: Hand-drawn diagram of a concave-up curve (U-shaped) with two plus signs above the endpoints, indicating that a positive second derivative represents an upward concavity and therefore a local minimum.

$ x = -1, \, x = 2 $ is Local Min.

\[ f''(-1) = 36 + 24 - 24 = 36 > 0 \]

Hand-drawn diagram of an upward-opening curve (concave-up) with two plus signs above it, confirming that a positive second derivative at x = -1 indicates upward concavity and a local minimum.

Visual Description: Hand-drawn diagram of an upward-opening curve (concave-up) with two plus signs above it, confirming that a positive second derivative at x = -1 indicates upward concavity and a local minimum.

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Example: $ y = \sin^2 x $ over $ [0, 2\pi] $

Apply second derivative test to determine Relative max/min.

\[ f'(x) = 2 \sin x \cos x = 0 \rightsquigarrow \sin x = 0 \quad \text{or} \quad \cos x = 0 \]

For $ \sin x = 0 $:

\[ 0, \pi, 2\pi \]

For $ \cos x = 0 $:

\[ \pi/2, 3\pi/2 \] \[ f''(x) = 2 \cos^2 x - 2 \sin^2 x \] \[ \left. \begin{aligned} f''(0) &= 2 \\ f''(\pi) &= 2 \\ f''(2\pi) &= 2 \end{aligned} \right\} > 0 \]

Mnemonic illustration for the second derivative test showing a happy face: two plus signs '+' serve as the eyes representing a positive second derivative (f'' > 0), and an upward-curving smile represents an upward concavity (concave up), indicating a local minimum.

Visual Description: Mnemonic illustration for the second derivative test showing a happy face: two plus signs '+' serve as the eyes representing a positive second derivative (f'' > 0), and an upward-curving smile represents an upward concavity (concave up), indicating a local minimum.

$ x = 0, x = \pi, x = 2\pi $, we have Local min.

\[ \left. \begin{aligned} f''(\pi/2) &= -2 \\ f''(3\pi/2) &= -2 \end{aligned} \right\} < 0 \]

Mnemonic illustration for the second derivative test showing a sad face: two minus signs '-' serve as the eyes representing a negative second derivative (f'' < 0), and a downward-curving frown represents a downward concavity (concave down), indicating a local maximum.

Visual Description: Mnemonic illustration for the second derivative test showing a sad face: two minus signs '-' serve as the eyes representing a negative second derivative (f'' < 0), and a downward-curving frown represents a downward concavity (concave down), indicating a local maximum.

$ x = \pi/2, x = 3\pi/2 $, we have Local Max.

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What happens when $ f'' = 0 $

eg: $ f(x) = x^3 \leadsto $

$ f'(x) = 3x^2 $
$ f''(x) = 6x $

$ f''(x) > 0 $ on $ (0, \infty) $ Concave up
$ f''(x) < 0 $ on $ (-\infty, 0) $ Concave down
$ f''(0) = 0 $

Cartesian coordinate plane depicting the graph of the cubic function f(x) = x^3. A blue curve starts in the third quadrant concave down, passes through the origin (0, 0) marked by a red point, and enters the first quadrant where it curves upwards (concave up). The origin marks the transition point between the two concavities.

Visual Description: Cartesian coordinate plane depicting the graph of the cubic function f(x) = x^3. A blue curve starts in the third quadrant concave down, passes through the origin (0, 0) marked by a red point, and enters the first quadrant where it curves upwards (concave up). The origin marks the transition point between the two concavities.

at $ x = 0 $, it changes from concave up to concave down.

$ x = 0 $ is called inflection points

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Example 1:

\[ f(x) = -x^3 \]

\[ f'(x) = -3x^2 < 0 \quad \text{for } x \ne 0 \]

\[ f''(x) = -6x \]

\[ f''(x) > 0 \quad \text{on } (-\infty, 0) \]

\[ f''(x) < 0 \quad \text{on } (0, \infty) \]

\[ f''(0) = 0 \]

Sketches illustrating concavity: an upper curve bending upwards labeled 'c. up' for concave up, and a lower curve bending downwards labeled 'c. down' for concave down.

Visual Description: Sketches illustrating concavity: an upper curve bending upwards labeled 'c. up' for concave up, and a lower curve bending downwards labeled 'c. down' for concave down.

Graph of the function f(x) = -x^3 on Cartesian coordinate axes. The cubic curve decreases monotonically, passing through the origin (0,0) where the tangent is horizontal. The curve is concave up on the interval (-inf, 0) and concave down on the interval (0, inf).

Visual Description: Graph of the function f(x) = -x^3 on Cartesian coordinate axes. The cubic curve decreases monotonically, passing through the origin (0,0) where the tangent is horizontal. The curve is concave up on the interval (-inf, 0) and concave down on the interval (0, inf).

at $x = 0$ concavity changes $\Rightarrow$ it's an inflection.

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e.g.

\[ f(x) = x^4 \]

\[ f'(x) = 4x^3 \]

\[ f''(x) = 12x^2 \]

\[ f''(x) > 0 \text{ for } x \neq 0 \implies \text{Concave up except at } x = 0 \]

\[ f''(0) = 0 \]

Cartesian coordinate plane showing the graph of the function f(x) = x^4 drawn in blue. The curve rests on the horizontal x-axis at the origin (0, 0), where it flattens, and extends symmetrically upward into both the first and second quadrants, remaining concave upward everywhere.

Visual Description: Cartesian coordinate plane showing the graph of the function f(x) = x^4 drawn in blue. The curve rests on the horizontal x-axis at the origin (0, 0), where it flattens, and extends symmetrically upward into both the first and second quadrants, remaining concave upward everywhere.

$ x = 0 $ is NOT an inflection because concavity does change at $ x = 0 $.

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eg: $f(x) = xe^x$, find inflection points and intervals where the function is concave up and concave down.

\[ f'(x) = e^x + xe^x \] \[ f''(x) = e^x + e^x + xe^x = e^x[2+x] \] \[ e^x > 0 \quad \text{for all } x \] \[ 2+x > 0 \quad \text{for } x > -2 \] \[ 2+x < 0 \quad \text{for } x < -2 \] \[ 2+x = 0 \quad \text{for } x = -2 \]

Sign chart on a horizontal number line testing the second derivative f''(x) around the critical value x = -2. To the left of -2, labeled f'', a series of minus signs indicates that f''(x) is negative, accompanied below by an inverted U-shaped curve showing the function is concave down. To the right of -2, labeled f'', a series of plus signs indicates that f''(x) is positive, accompanied below by an upright U-shaped curve showing the function is concave up.

Visual Description: Sign chart on a horizontal number line testing the second derivative f''(x) around the critical value x = -2. To the left of -2, labeled f'', a series of minus signs indicates that f''(x) is negative, accompanied below by an inverted U-shaped curve showing the function is concave down. To the right of -2, labeled f'', a series of plus signs indicates that f''(x) is positive, accompanied below by an upright U-shaped curve showing the function is concave up.

$(-\infty, -2)$ c. down

$(-2, \infty)$ is c. up,

$x = -2$ is inflection pt.

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Example:

$ f(x) = x^4 - 4x^3 + 6x^2 $

Find inflection points, intervals of concave up and concave down!

\[ f'(x) = 4x^3 - 12x^2 + 12x \]

\[ f''(x) = 12x^2 - 24x + 12 = 12(x^2 - 2x + 1) = 12(x - 1)^2 \]

\[ \begin{aligned} f''(x) > 0 & \quad x \neq 1 \\ = 0 & \quad x = 1 \end{aligned} \]

Sign chart on a horizontal number line centered around x = 1. A vertical tick mark labels the point 1. To the left of 1, four plus (+) signs are marked above the line, with an upward-opening curve underneath signifying concave up. To the right of 1, three plus (+) signs are marked above the line, also with an upward-opening curve underneath signifying concave up. Since the sign of the second derivative remains positive on both sides of x = 1, the concavity does not change.

Visual Description: Sign chart on a horizontal number line centered around x = 1. A vertical tick mark labels the point 1. To the left of 1, four plus (+) signs are marked above the line, with an upward-opening curve underneath signifying concave up. To the right of 1, three plus (+) signs are marked above the line, also with an upward-opening curve underneath signifying concave up. Since the sign of the second derivative remains positive on both sides of x = 1, the concavity does not change.

No inflection points

interval of c. up $ (-\infty, 1) \cup (1, \infty) $

Lesson 23 (10/23/26)

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Lesson 23 (10/23/26)

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Review: First Derivative Test

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eg 1

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What does \( f''(x) \) tell us?

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2. \( f'' < 0 \sim \frac{d}{dx}(f'(x)) < 0 \Rightarrow f'(x) \text{ is } \downarrow \)

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Second Derivative Test:

Relative max at \( x = c \):

Relative min at \( x = c \):

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e.g.:

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Example: \( y = \sin^2 x \) over \( [0, 2\pi] \)

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What happens when \( f'' = 0 \)

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Example 1:

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e.g.

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Example: