Lecture: Section 4.3 Derivatives and the Shape of a Graph

In this section, we'll delve deeper into how the first and second derivatives of a function, $f'(x)$ and $f''(x)$, reveal crucial information about the shape and behavior of the original function's graph, $y=f(x)$. We'll learn how to determine where a function is rising or falling, identify its peaks and valleys, and understand its curvature.

Topic 1: Increasing and Decreasing Functions 📈📉

We intuitively know what it means for a graph to go "uphill" (increasing) or "downhill" (decreasing) as we move from left to right. The first derivative gives us a precise way to determine this.

Definitions of Increasing and Decreasing

A function $f$ is called:

Increasing on an interval $I$ if $f(x_1) < f(x_2)$ whenever $x_1 < x_2$ in $I$.
Decreasing on an interval $I$ if $f(x_1) > f(x_2)$ whenever $x_1 < x_2$ in $I$.

Think about the slope of the tangent line. When the function is increasing, the tangent lines have positive slopes. When the function is decreasing, the tangent lines have negative slopes. This leads to a powerful test:

Increasing/Decreasing (I/D) Test

If $f'(x) > 0$ on an interval, then $f$ is increasing on that interval.
If $f'(x) < 0$ on an interval, then $f$ is decreasing on that interval.

To apply this test, we first find the critical numbers of $f$ (where $f'(x)=0$ or $f'(x)$ does not exist). These numbers divide the domain into intervals. We then test the sign of $f'(x)$ in each interval.

Move the slider a to see the relationship between the sign of the derivative and where f is increasing/decreasing

Example 1: Finding Intervals of Increase/Decrease (Polynomial)

Find where the function $f(x) = x^3 - 6x^2 + 5$ is increasing and decreasing.

Solution:

1. Find the derivative:

$$f'(x) = 3x^2 - 12x$$

2. Find critical numbers: Set $f'(x) = 0$.

$$3x^2 - 12x = 0 \implies 3x(x-4) = 0$$

The critical numbers are $x=0$ and $x=4$. (Since $f'(x)$ is a polynomial, it exists everywhere).

3. Create intervals and test signs: The critical numbers divide the number line into $(-\infty, 0)$, $(0, 4)$, and $(4, \infty)$. We pick a test value in each interval and check the sign of $f'(x)$.

Interval $(-\infty, 0)$: Test $x=-1$. $f'(-1) = 3(-1)^2 - 12(-1) = 3 + 12 = 15 > 0$.
Interval $(0, 4)$: Test $x=1$. $f'(1) = 3(1)^2 - 12(1) = 3 - 12 = -9 < 0$.
Interval $(4, \infty)$: Test $x=5$. $f'(5) = 3(5)^2 - 12(5) = 75 - 60 = 15 > 0$.

4. Conclusion:

$f$ is increasing on $(-\infty, 0)$ and $(4, \infty)$ because $f'(x) > 0$.
$f$ is decreasing on $(0, 4)$ because $f'(x) < 0$.

Move the slider 'a' to see how the critical points divide the interval into where the function is increasing and decreasing.

Example 2: Finding Intervals of Increase/Decrease (Rational Function)

Find where the function $f(x) = \frac{x^2}{x-1}$ is increasing and decreasing.

Solution:

1. Note the domain: The function is undefined at $x=1$. Domain: $(-\infty, 1) \cup (1, \infty)$.

2. Find the derivative (Quotient Rule):

$$f'(x) = \frac{(x-1)(2x) - (x^2)(1)}{(x-1)^2} = \frac{2x^2 - 2x - x^2}{(x-1)^2} = \frac{x^2 - 2x}{(x-1)^2} = \frac{x(x-2)}{(x-1)^2}$$

3. Find critical numbers:

Set $f'(x) = 0$: $\frac{x(x-2)}{(x-1)^2} = 0 \implies x(x-2) = 0$. So $x=0$ and $x=2$ are critical numbers.
Find where $f'(x)$ DNE: The derivative is undefined at $x=1$, but $x=1$ is not in the domain of $f$. So it's not a critical number, but we must use it to define our intervals.

4. Create intervals and test signs: The points $x=0, x=1, x=2$ divide the domain into $(-\infty, 0)$, $(0, 1)$, $(1, 2)$, and $(2, \infty)$. Notice the denominator $(x-1)^2$ is always positive where defined.

Interval $(-\infty, 0)$: Test $x=-1$. $f'(-1) = \frac{(-1)(-1-2)}{(-1-1)^2} = \frac{(-1)(-3)}{(-2)^2} = \frac{3}{4} > 0$.
Interval $(0, 1)$: Test $x=0.5$. $f'(0.5) = \frac{(0.5)(0.5-2)}{(0.5-1)^2} = \frac{(0.5)(-1.5)}{(-0.5)^2} = \frac{-0.75}{0.25} < 0$.
Interval $(1, 2)$: Test $x=1.5$. $f'(1.5) = \frac{(1.5)(1.5-2)}{(1.5-1)^2} = \frac{(1.5)(-0.5)}{(0.5)^2} = \frac{-0.75}{0.25} < 0$.
Interval $(2, \infty)$: Test $x=3$. $f'(3) = \frac{(3)(3-2)}{(3-1)^2} = \frac{(3)(1)}{(2)^2} = \frac{3}{4} > 0$.

5. Conclusion:

$f$ is increasing on $(-\infty, 0)$ and $(2, \infty)$ because $f'(x) > 0$.
$f$ is decreasing on $(0, 1)$ and $(1, 2)$ because $f'(x) < 0$.

Check Your Understanding 🤔

Find the intervals on which $f(x) = x e^{-x}$ is increasing or decreasing.

Solution:

1. Find derivative (Product Rule):

$$f'(x) = (1)e^{-x} + x(-e^{-x}) = e^{-x}(1-x)$$

2. Find critical numbers: Set $f'(x)=0$. Since $e^{-x}$ is never zero, we only need $1-x=0$, which gives $x=1$. The derivative exists everywhere.

3. Create intervals and test signs: $(-\infty, 1)$ and $(1, \infty)$.

Interval $(-\infty, 1)$: Test $x=0$. $f'(0) = e^0(1-0) = 1 > 0$.
Interval $(1, \infty)$: Test $x=2$. $f'(2) = e^{-2}(1-2) = -e^{-2} < 0$.

4. Conclusion:

$f$ is increasing on $(-\infty, 1)$.
$f$ is decreasing on $(1, \infty)$.

Topic 2: The First Derivative Test for Local Extrema ⛰️

The points where a function changes from increasing to decreasing (or vice versa) are often local maximums or minimums (collectively called local extrema). The sign analysis of $f'(x)$ around its critical numbers tells us exactly what's happening.

The First Derivative Test

Suppose $c$ is a critical number of a continuous function $f$.

If $f'$ changes from positive to negative at $c$, then $f$ has a local maximum at $c$. (Function goes from increasing to decreasing).
If $f'$ changes from negative to positive at $c$, then $f$ has a local minimum at $c$. (Function goes from decreasing to increasing).
If $f'$ does not change sign at $c$ (e.g., positive on both sides or negative on both sides), then $f$ has no local maximum or minimum at $c$. (Like at $x=0$ for $f(x)=x^3$).

interval	$12x$	$x-2$	$x+1$	$f'(x)$	$f$
$x < -1$ (e.g., $x=-2$)	$-$	$-$	$-$	$-$	Decreasing on $(-\infty, -1)$
$-1 < x < 0$ (e.g., $x=-0.5$)	$-$	$-$	$+$	$+$	Increasing on $(-1, 0)$
$0 < x < 2$ (e.g., $x=1$)	$+$	$-$	$+$	$-$	Decreasing on $(0, 2)$
$x > 2$ (e.g., $x=3$)	$+$	$+$	$+$	$+$	Increasing on $(2, \infty)$

Essentially, we use the same sign analysis from the Increasing/Decreasing Test. We just add a conclusion about local extrema based on how the sign changes at each critical number.

Example 1: Applying the First Derivative Test (Polynomial)

Find the local maximum and minimum values of $f(x) = x^3 - 6x^2 + 5$ (from Topic 1, Example 1).

Solution:

From the previous example, we know:

$f'(x) = 3x(x-4)$.
Critical numbers are $x=0$ and $x=4$.
Sign analysis of $f'$:
$(-\infty, 0)$: $f' > 0$ (Increasing)
$(0, 4)$: $f' < 0$ (Decreasing)
$(4, \infty)$: $f' > 0$ (Increasing)

Apply the test:

At $x=0$: $f'$ changes from positive to negative (+ to -). Therefore, $f$ has a local maximum at $x=0$. The local maximum value is $f(0) = 0^3 - 6(0)^2 + 5 = 5$.
At $x=4$: $f'$ changes from negative to positive (- to +). Therefore, $f$ has a local minimum at $x=4$. The local minimum value is $f(4) = 4^3 - 6(4)^2 + 5 = 64 - 96 + 5 = -27$.

Conclusion: Local maximum value is 5 at $x=0$. Local minimum value is -27 at $x=4$.

Example 2: Applying the First Derivative Test (Rational Function)

Find the local maximum and minimum values of $f(x) = \frac{x^2}{x-1}$ (from Topic 1, Example 2).

Solution:

From the previous example, we know:

$f'(x) = \frac{x(x-2)}{(x-1)^2}$.
Critical numbers are $x=0$ and $x=2$. ($x=1$ is not in the domain).
Sign analysis of $f'$:
$(-\infty, 0)$: $f' > 0$ (Increasing)
$(0, 1)$: $f' < 0$ (Decreasing)
$(1, 2)$: $f' < 0$ (Decreasing)
$(2, \infty)$: $f' > 0$ (Increasing)

Apply the test:

At $x=0$: $f'$ changes from positive to negative (+ to -). Therefore, $f$ has a local maximum at $x=0$. The value is $f(0) = \frac{0^2}{0-1} = 0$.
At $x=1$: This is not a critical number (it's a vertical asymptote), so the test doesn't apply here. There's no local extremum at $x=1$.
At $x=2$: $f'$ changes from negative to positive (- to +). Therefore, $f$ has a local minimum at $x=2$. The value is $f(2) = \frac{2^2}{2-1} = \frac{4}{1} = 4$.

Conclusion: Local maximum value is 0 at $x=0$. Local minimum value is 4 at $x=2$.

Check Your Understanding 🤔

Find the local maximum and minimum values of $f(x) = x e^{-x}$ (from the previous CYU).

Solution:

From the previous CYU, we know:

$f'(x) = e^{-x}(1-x)$.
Critical number is $x=1$.
Sign analysis of $f'$:
$(-\infty, 1)$: $f' > 0$ (Increasing)
$(1, \infty)$: $f' < 0$ (Decreasing)

Apply the test:

At $x=1$: $f'$ changes from positive to negative (+ to -). Therefore, $f$ has a local maximum at $x=1$. The value is $f(1) = 1 \cdot e^{-1} = 1/e$.

Conclusion: Local maximum value is $1/e$ at $x=1$. There is no local minimum.

Topic 3: Concavity and Inflection Points ☕🥄

While the first derivative tells us if the function is increasing or decreasing, the second derivative tells us about the *way* it's increasing or decreasing – its curvature or "concavity."

Definition of Concavity

If the graph of $f$ lies above all of its tangents on an interval $I$, it is called concave upward (CU) on $I$. (Like a cup holding water ☕).
If the graph of $f$ lies below all of its tangents on an interval $I$, it is called concave downward (CD) on $I$. (Like a spoon spilling water 🥄).

How does $f''$ relate? If $f'' > 0$, then $f'$ is increasing. This means the slopes of the tangent lines are getting larger (less negative, or more positive), causing the graph to bend upward (CU). If $f'' < 0$, then $f'$ is decreasing, the slopes are getting smaller, and the graph bends downward (CD).

Concavity Test

If $f''(x) > 0$ for all $x$ in $I$, then the graph of $f$ is concave upward on $I$.
If $f''(x) < 0$ for all $x$ in $I$, then the graph of $f$ is concave downward on $I$.

A point where the concavity changes is called an inflection point.

Definition of Inflection Point

A point $P$ on a curve $y=f(x)$ is called an inflection point if $f$ is continuous there and the curve changes from concave upward to concave downward, or vice versa, at $P$. These often occur where $f''(x)=0$ or $f''(x)$ DNE.

Example 1: Finding Concavity and Inflection Points (Polynomial)

Determine the intervals of concavity and find the inflection points for $f(x) = x^4 - 6x^2$.

Solution:

1. Find first and second derivatives:

$$f'(x) = 4x^3 - 12x$$ $$f''(x) = 12x^2 - 12$$

2. Find where $f''(x)=0$ or DNE: Set $f''(x)=0$.

$$12x^2 - 12 = 0 \implies 12(x^2-1) = 0 \implies 12(x-1)(x+1) = 0$$

Possible inflection points are at $x=1$ and $x=-1$. ($f''(x)$ exists everywhere).

3. Create intervals and test signs of $f''(x)$: The points $x=-1, x=1$ divide the number line into $(-\infty, -1)$, $(-1, 1)$, and $(1, \infty)$.

Interval $(-\infty, -1)$: Test $x=-2$. $f''(-2) = 12(-2)^2 - 12 = 12(4) - 12 = 36 > 0$.
Interval $(-1, 1)$: Test $x=0$. $f''(0) = 12(0)^2 - 12 = -12 < 0$.
Interval $(1, \infty)$: Test $x=2$. $f''(2) = 12(2)^2 - 12 = 12(4) - 12 = 36 > 0$.

4. Determine concavity:

Concave Upward (CU) on $(-\infty, -1)$ and $(1, \infty)$ because $f''(x) > 0$.
Concave Downward (CD) on $(-1, 1)$ because $f''(x) < 0$.

5. Identify inflection points: Concavity changes at both $x=-1$ and $x=1$. We find the y-coordinates:

$f(-1) = (-1)^4 - 6(-1)^2 = 1 - 6 = -5$. Point: $(-1, -5)$.
$f(1) = (1)^4 - 6(1)^2 = 1 - 6 = -5$. Point: $(1, -5)$.

Conclusion: CU on $(-\infty, -1) \cup (1, \infty)$; CD on $(-1, 1)$. Inflection points at $(-1, -5)$ and $(1, -5)$.

Example 2: Concavity for a Function with Fractional Exponent

Discuss the concavity and find inflection points for $f(x) = x^{1/3}(x+4)$.

Solution:

First, simplify $f(x) = x^{4/3} + 4x^{1/3}$.

1. Find derivatives:

$$f'(x) = \frac{4}{3}x^{1/3} + \frac{4}{3}x^{-2/3} = \frac{4}{3}x^{-2/3}(x+1)$$ $$f''(x) = \frac{4}{9}x^{-2/3} + \frac{4}{3}\left(-\frac{2}{3}\right)x^{-5/3} = \frac{4}{9}x^{-2/3} - \frac{8}{9}x^{-5/3}$$

Simplify $f''(x)$ by factoring out $\frac{4}{9}x^{-5/3}$ (the lowest power):

$$f''(x) = \frac{4}{9}x^{-5/3}(x - 2) = \frac{4(x-2)}{9x^{5/3}}$$

2. Find where $f''(x)=0$ or DNE:

$f''(x)=0$ when the numerator is zero: $4(x-2)=0 \implies x=2$.
$f''(x)$ DNE when the denominator is zero: $9x^{5/3}=0 \implies x=0$.

Possible inflection points occur at $x=0$ and $x=2$.

3. Create intervals and test signs of $f''(x)$: $(-\infty, 0)$, $(0, 2)$, $(2, \infty)$.

Interval $(-\infty, 0)$: Test $x=-1$. $f''(-1) = \frac{4(-1-2)}{9(-1)^{5/3}} = \frac{4(-3)}{9(-1)} = \frac{-12}{-9} > 0$.
Interval $(0, 2)$: Test $x=1$. $f''(1) = \frac{4(1-2)}{9(1)^{5/3}} = \frac{4(-1)}{9(1)} = \frac{-4}{9} < 0$.
Interval $(2, \infty)$: Test $x=8$. $f''(8) = \frac{4(8-2)}{9(8)^{5/3}} = \frac{4(6)}{9(32)} = \frac{24}{288} > 0$.

4. Determine concavity:

CU on $(-\infty, 0)$ and $(2, \infty)$ where $f''(x)>0$.
CD on $(0, 2)$ where $f''(x)<0$.

5. Identify inflection points: Concavity changes at both $x=0$ and $x=2$. $f(x)$ is continuous at both points.

$f(0) = 0^{1/3}(0+4) = 0$. Point: $(0, 0)$.
$f(2) = 2^{1/3}(2+4) = 6\sqrt[3]{2}$. Point: $(2, 6\sqrt[3]{2})$.

Conclusion: CU on $(-\infty, 0) \cup (2, \infty)$; CD on $(0, 2)$. Inflection points at $(0, 0)$ and $(2, 6\sqrt[3]{2})$.

Check Your Understanding 🤔

Find the intervals of concavity and the inflection points of $f(x) = \ln(x^2+1)$.

Solution:

1. Find derivatives:

$$f'(x) = \frac{1}{x^2+1} \cdot (2x) = \frac{2x}{x^2+1}$$ $$f''(x) = \frac{(x^2+1)(2) - (2x)(2x)}{(x^2+1)^2} = \frac{2x^2+2 - 4x^2}{(x^2+1)^2} = \frac{2-2x^2}{(x^2+1)^2} = \frac{2(1-x^2)}{(x^2+1)^2} = \frac{2(1-x)(1+x)}{(x^2+1)^2}$$

2. Find where $f''(x)=0$ or DNE: Set $f''(x)=0$. The denominator $(x^2+1)^2$ is always positive. The numerator is zero when $1-x^2=0$, so $x=1$ and $x=-1$. $f''(x)$ exists everywhere.

3. Create intervals and test signs of $f''(x)$: $(-\infty, -1)$, $(-1, 1)$, $(1, \infty)$.

Interval $(-\infty, -1)$: Test $x=-2$. $f''(-2) = \frac{2(1-(-2)^2)}{((-2)^2+1)^2} = \frac{2(1-4)}{(4+1)^2} = \frac{-6}{25} < 0$.
Interval $(-1, 1)$: Test $x=0$. $f''(0) = \frac{2(1-0^2)}{(0^2+1)^2} = \frac{2}{1} > 0$.
Interval $(1, \infty)$: Test $x=2$. $f''(2) = \frac{2(1-(2)^2)}{((2)^2+1)^2} = \frac{2(1-4)}{(4+1)^2} = \frac{-6}{25} < 0$.

4. Determine concavity:

Concave Downward (CD) on $(-\infty, -1)$ and $(1, \infty)$.
Concave Upward (CU) on $(-1, 1)$.

5. Identify inflection points: Concavity changes at $x=-1$ and $x=1$.

$f(-1) = \ln((-1)^2+1) = \ln(2)$. Point: $(-1, \ln 2)$.
$f(1) = \ln((1)^2+1) = \ln(2)$. Point: $(1, \ln 2)$.

Conclusion: CD on $(-\infty, -1) \cup (1, \infty)$; CU on $(-1, 1)$. Inflection points at $(-1, \ln 2)$ and $(1, \ln 2)$.

Topic 4: The Second Derivative Test for Local Extrema 🤔

There's another way to classify local extrema, specifically at points where $f'(c)=0$. Instead of checking the sign of $f'$ on either side of $c$, we can sometimes just check the sign of $f''(c)$ *at* $c$.

The Second Derivative Test

Suppose $f''$ is continuous near $c$.

If $f'(c) = 0$ and $f''(c) > 0$, then $f$ has a local minimum at $c$. (Think: $f'(c)=0$ means flat tangent, $f''(c)>0$ means concave up like a cup $\implies$ bottom of the cup).
If $f'(c) = 0$ and $f''(c) < 0$, then $f$ has a local maximum at $c$. (Think: $f'(c)=0$ means flat tangent, $f''(c)<0$ means concave down like a frown $\implies$ top of the frown).
If $f'(c) = 0$ and $f''(c) = 0$, the test is inconclusive. $f$ might have a local max, local min, or neither at $c$. (You must use the First Derivative Test in this case).

This test is often quicker than the First Derivative Test if the second derivative is easy to compute and evaluate. However, its major drawback is that it fails when $f''(c)=0$.

Example 1: Using the Second Derivative Test

Use the Second Derivative Test to find the local extrema of $f(x) = x^3 - 6x^2 + 5$.

Solution:

1. Find $f'(x)$ and critical numbers (where $f'=0$):

$$f'(x) = 3x^2 - 12x = 3x(x-4)$$

Critical numbers (where $f'=0$) are $x=0$ and $x=4$.

2. Find $f''(x)$:

$$f''(x) = 6x - 12$$

3. Evaluate $f''(x)$ at each critical number:

At $x=0$: $f''(0) = 6(0) - 12 = -12$. Since $f''(0) < 0$, the Second Derivative Test indicates a local maximum at $x=0$. The value is $f(0)=5$.
At $x=4$: $f''(4) = 6(4) - 12 = 24 - 12 = 12$. Since $f''(4) > 0$, the Second Derivative Test indicates a local minimum at $x=4$. The value is $f(4)=-27$.

Conclusion: Local max at $(0, 5)$, local min at $(4, -27)$. (This matches the result from the First Derivative Test example).

Example 2: Inconclusive Second Derivative Test

Try to use the Second Derivative Test to classify the critical point of $f(x) = x^4$.

Solution:

1. Find $f'(x)$ and critical numbers (where $f'=0$):

$$f'(x) = 4x^3$$

Set $f'(x)=0 \implies 4x^3=0 \implies x=0$. The only critical number is $x=0$.

2. Find $f''(x)$:

$$f''(x) = 12x^2$$

3. Evaluate $f''(x)$ at the critical number:

$$f''(0) = 12(0)^2 = 0$$

Conclusion: Since $f''(0)=0$, the Second Derivative Test is inconclusive. We cannot determine if $x=0$ corresponds to a local max, min, or neither using this test.

(Note: If we used the First Derivative Test: $f'(x)=4x^3$ is negative for $x<0$ and positive for $x>0$. Since $f'$ changes from - to + at $x=0$, there is a local minimum there. The Second Derivative Test failed, but the First Derivative Test worked.)

Check Your Understanding 🤔

Use the Second Derivative Test to find the local extrema of $f(x) = x e^{-x}$.

Solution:

1. Find $f'(x)$ and critical numbers (where $f'=0$):

From previous work, $f'(x) = e^{-x}(1-x)$. The only critical number where $f'=0$ is $x=1$.

2. Find $f''(x)$ (using product rule on $f'$):

$$f''(x) = (e^{-x})(-1) + (1-x)(-e^{-x})$$ $$f''(x) = -e^{-x} - e^{-x} + xe^{-x} = e^{-x}(x-2)$$

3. Evaluate $f''(x)$ at the critical number $x=1$:

$$f''(1) = e^{-1}(1-2) = e^{-1}(-1) = -1/e$$

Conclusion: Since $f'(1)=0$ and $f''(1) < 0$, the Second Derivative Test indicates a local maximum at $x=1$. The value is $f(1)=1/e$.

Final Practice Problems

Practice Problem #1 (Comprehensive Analysis)

Let $f(x) = 2x^3 - 3x^2 - 12x + 1$. Find:

Intervals where $f$ is increasing/decreasing.
Local maximum and minimum values.
Intervals where $f$ is concave upward/downward.
Inflection points.

Solution:

a) & b) Increase/Decrease & Local Extrema (Use $f'$):

$$f'(x) = 6x^2 - 6x - 12 = 6(x^2 - x - 2) = 6(x-2)(x+1)$$

Critical numbers: $x=2, x=-1$.

Sign analysis for $f'$:

$(-\infty, -1)$: Test $x=-2$. $f'(-2) = 6(-4)(-1) > 0$. Increasing.
$(-1, 2)$: Test $x=0$. $f'(0) = 6(-2)(1) < 0$. Decreasing.
$(2, \infty)$: Test $x=3$. $f'(3) = 6(1)(4) > 0$. Increasing.

Increasing on $(-\infty, -1) \cup (2, \infty)$. Decreasing on $(-1, 2)$.

First Derivative Test:

Local Max at $x=-1$ (+ to - change). $f(-1) = 2(-1)^3 - 3(-1)^2 - 12(-1) + 1 = -2 - 3 + 12 + 1 = 8$.
Local Min at $x=2$ (- to + change). $f(2) = 2(2)^3 - 3(2)^2 - 12(2) + 1 = 16 - 12 - 24 + 1 = -19$.

Local max value: 8 at $x=-1$. Local min value: -19 at $x=2$.

c) & d) Concavity & Inflection Points (Use $f''$):

$$f''(x) = 12x - 6 = 6(2x - 1)$$

Possible inflection point: $f''(x)=0 \implies 2x-1=0 \implies x=1/2$.

Sign analysis for $f''$:

$(-\infty, 1/2)$: Test $x=0$. $f''(0) = -6 < 0$. Concave Downward (CD).
$(1/2, \infty)$: Test $x=1$. $f''(1) = 12-6 = 6 > 0$. Concave Upward (CU).

Concave Downward on $(-\infty, 1/2)$. Concave Upward on $(1/2, \infty)$.

Inflection Point: Concavity changes at $x=1/2$. $f(1/2) = 2(1/8) - 3(1/4) - 12(1/2) + 1 = 1/4 - 3/4 - 6 + 1 = -2/4 - 5 = -1/2 - 5 = -11/2$.

Inflection point at $(1/2, -11/2)$.

Practice Problem #2 (Given $f'$)

The derivative of a function $f$ is given by $f'(x) = (x+3)x^2(x-2)$.

Find the intervals where $f$ is increasing or decreasing.
Find the x-coordinates of the local extrema of $f$.

Solution:

We analyze the sign of $f'(x) = (x+3)x^2(x-2)$.

Critical numbers: $f'(x)=0$ when $x=-3, x=0, x=2$.

Sign analysis for $f'$: Note that $x^2$ is always $\ge 0$.

$(-\infty, -3)$: Test $x=-4$. $f'(-4) = (-1)(-4)^2(-6) = (-)(+)(-) > 0$. Increasing.
$(-3, 0)$: Test $x=-1$. $f'(-1) = (2)(-1)^2(-3) = (+)(+)(-) < 0$. Decreasing.
$(0, 2)$: Test $x=1$. $f'(1) = (4)(1)^2(-1) = (+)(+)(-) < 0$. Decreasing.
$(2, \infty)$: Test $x=3$. $f'(3) = (6)(3)^2(1) = (+)(+)(+) > 0$. Increasing.

a) Increasing on $(-\infty, -3) \cup (2, \infty)$. Decreasing on $(-3, 0) \cup (0, 2)$ (or just $(-3, 2)$).

b) First Derivative Test:

At $x=-3$: $f'$ changes + to -. Local maximum.
At $x=0$: $f'$ does not change sign (- to -). No local extremum.
At $x=2$: $f'$ changes - to +. Local minimum.

Local maximum at $x=-3$. Local minimum at $x=2$.

Practice Problem #3 (Interpreting Graph of $f'$)

The graph of the derivative $f'$ of a continuous function $f$ is shown below. Assume $f$ is continuous everywhere.

On what intervals is $f$ increasing or decreasing?
At what values of $x$ does $f$ have a local maximum or minimum?
On what intervals is $f$ concave upward or downward?
State the $x$-coordinate(s) of the inflection point(s).

Solution:

We analyze the given graph of $f'$ to deduce properties of $f$.

a) Increase/Decrease: $f$ increases where $f' > 0$ and decreases where $f' < 0$.

From the graph, $f'(x) > 0$ (graph above x-axis) on $(-\infty, 1) \cup (5, 7)$. $\implies f$ is increasing.
From the graph, $f'(x) < 0$ (graph below x-axis) on $(1, 5) \cup (7, \infty)$. $\implies f$ is decreasing.

$f$ is increasing on $(-\infty, 1)$ and $(5, 7)$. $f$ is decreasing on $(1, 5)$ and $(7, \infty)$.

b) Local Extrema: Occur where $f'$ changes sign (crosses the x-axis).

At $x=1$: $f'$ changes from + to -. $\implies$ Local maximum.
At $x=5$: $f'$ changes from - to +. $\implies$ Local minimum.
At $x=7$: $f'$ changes from + to -. $\implies$ Local maximum.

Local maxima at $x=1$ and $x=7$. Local minimum at $x=5$.

c) Concavity: $f$ is CU where $f'' > 0$ (i.e., where $f'$ is increasing) and CD where $f'' < 0$ (i.e., where $f'$ is decreasing).

From the graph, $f'(x)$ is increasing (graph slope is positive) on approx. $(-\infty, 2.57)$ and $(6.10, \infty)$. $\implies f$ is CU.
From the graph, $f'(x)$ is decreasing (graph slope is negative) on approx. $(2.57, 6.10)$. $\implies f$ is CD.

(Using $f''(x) = -3x^2 + 26x - 47$, the roots are $x = \frac{13 \mp 2\sqrt{7}}{3}$.)

Concave Upward on $\left(-\infty, \frac{13 - 2\sqrt{7}}{3}\right) \cup \left(\frac{13 + 2\sqrt{7}}{3}, \infty\right)$. Concave Downward on $\left(\frac{13 - 2\sqrt{7}}{3}, \frac{13 + 2\sqrt{7}}{3}\right)$.

d) Inflection Points: Occur where concavity changes (i.e., where $f'$ changes direction, at its local extrema).

From the graph, $f'$ has local extrema (peaks/valleys) at the points where $f''(x)=0$.

Inflection points at $x = \frac{13 - 2\sqrt{7}}{3} \approx 2.57$ and $x = \frac{13 + 2\sqrt{7}}{3} \approx 6.10$.

Lecture Conclusion ✅

Today, we established the fundamental connections between a function and its first and second derivatives regarding the shape of its graph. The sign of $f'$ tells us whether the function is rising or falling, and the sign of $f''$ tells us about its concavity.

Key Takeaways

Increasing/Decreasing Test: $f' > 0 \implies$ Increasing, $f' < 0 \implies$ Decreasing. Find intervals using critical numbers.
First Derivative Test: Classifies local extrema based on sign changes in $f'$ at critical numbers (+ to - is local max, - to + is local min).
Concavity Test: $f'' > 0 \implies$ Concave Up (CU), $f'' < 0 \implies$ Concave Down (CD). Find intervals using points where $f''=0$ or DNE.
Inflection Points: Occur where concavity changes (often where $f''=0$ or DNE).
Second Derivative Test: An alternative way to classify local extrema where $f'(c)=0$: If $f''(c)>0$, local min; if $f''(c)<0$, local max; if $f''(c)=0$, inconclusive.

These tests are the building blocks for accurately sketching the graph of a function, which we will explore further in Section 4.5.