The Central Idea
The derivative
\[ f'(x) \]tells us the local slope of \(f\).
Its sign determines whether the graph is rising or falling:
\[ \boxed{ f'(x)>0 \Rightarrow f\text{ is increasing} } \] \[ \boxed{ f'(x)<0 \Rightarrow f\text{ is decreasing} }. \]The second derivative
\[ f''(x) \]tells us how those slopes are changing and therefore describes concavity.
Increasing Functions
A function is increasing on an interval if larger inputs correspond to larger outputs throughout that interval.
In derivative language, if
\[ f'(x)>0 \]throughout an interval, then \(f\) is increasing there.
Geometrically, the tangent lines have positive slope.
Decreasing Functions
A function is decreasing on an interval if larger inputs correspond to smaller outputs throughout that interval.
If
\[ f'(x)<0, \]then the tangent lines have negative slope and the graph is decreasing.
Horizontal Tangents
If
\[ f'(c)=0, \]the tangent line at \(x=c\) is horizontal.
Such an input is a critical number and may correspond to a local maximum, local minimum, or a flat point that is neither.
The derivative value alone does not tell us which one occurs.
Critical Numbers Divide the Domain into Test Intervals
Critical numbers provide natural boundaries for studying the sign of \(f'(x)\).
Suppose the critical numbers are
\[ c_1We can test the sign of \(f'(x)\) on each interval.
First-Derivative Sign Charts
A sign chart records whether the derivative is positive, negative, or zero.
For example,
| Interval or point | Sign of \(f'(x)\) | Behavior of \(f\) |
|---|---|---|
| \((-\infty,1)\) | Positive | Increasing |
| \(x=1\) | \(0\) | Horizontal tangent |
| \((1,\infty)\) | Negative | Decreasing |
The change from positive to negative indicates a local maximum at \(x=1\).
What Is a Local Maximum?
A function has a local maximum at \(x=c\) if \(f(c)\) is greater than or equal to nearby function values.
Informally, the graph reaches a nearby high point.
A common derivative pattern is
\[ f'(x)>0 \]before \(c\), followed by
\[ f'(x)<0 \]after \(c\).
\[ \boxed{ +\ \longrightarrow\ - \quad\Rightarrow\quad \text{local maximum} }. \]What Is a Local Minimum?
A function has a local minimum at \(x=c\) if \(f(c)\) is less than or equal to nearby function values.
A common derivative pattern is
\[ f'(x)<0 \]before \(c\), followed by
\[ f'(x)>0 \]after \(c\).
\[ \boxed{ -\ \longrightarrow\ + \quad\Rightarrow\quad \text{local minimum} }. \]The First Derivative Test
The first derivative test classifies critical numbers by examining how the sign of \(f'(x)\) changes around them.
- \(+\) to \(-\): local maximum.
- \(-\) to \(+\): local minimum.
- No sign change: neither a local maximum nor a local minimum.
This test is especially useful because it follows directly from increasing and decreasing behavior.
Example: First Derivative Test
Let
\[ f(x)=x^3-3x. \]Differentiate:
\[ f'(x)=3x^2-3. \]Factor:
\[ f'(x)=3(x-1)(x+1). \]The critical numbers are
\[ \boxed{x=-1,\ 1}. \]Building the Sign Chart
The critical numbers divide the line into
\[ (-\infty,-1), \qquad (-1,1), \qquad (1,\infty). \]Test one value in each interval.
| Interval | Test value | Sign of \(f'(x)\) | Behavior |
|---|---|---|---|
| \((-\infty,-1)\) | \(-2\) | Positive | Increasing |
| \((-1,1)\) | \(0\) | Negative | Decreasing |
| \((1,\infty)\) | \(2\) | Positive | Increasing |
Classifying the Critical Numbers
At \(x=-1\), the derivative changes from positive to negative:
\[ +\to-. \]Therefore, \(x=-1\) gives a local maximum.
At \(x=1\), the derivative changes from negative to positive:
\[ -\to+. \]Therefore, \(x=1\) gives a local minimum.
Finding the Local Extreme Points
Continue with
\[ f(x)=x^3-3x. \]Evaluate:
\[ f(-1)=-1+3=2, \] \[ f(1)=1-3=-2. \]Therefore, the local maximum point is
\[ \boxed{(-1,2)} \]and the local minimum point is
\[ \boxed{(1,-2)}. \]A Critical Point with No Extremum
Consider
\[ f(x)=x^3. \]Then
\[ f'(x)=3x^2. \]The only critical number is
\[ x=0. \]But
\[ f'(x)>0 \]on both sides of \(0\).
Thus, the function is increasing before and after the critical number, so the origin is neither a local maximum nor a local minimum.
Local versus Absolute Extrema
A local extremum is highest or lowest relative to nearby points.
An absolute extremum is highest or lowest over the entire domain or over a specified interval.
A local maximum need not be the absolute maximum, and a local minimum need not be the absolute minimum.
Absolute Maximum
A function has an absolute maximum at \(x=c\) on a set if
\[ f(c)\ge f(x) \]for every \(x\) in that set.
The value
\[ f(c) \]is the greatest function value under consideration.
Absolute Minimum
A function has an absolute minimum at \(x=c\) on a set if
\[ f(c)\le f(x) \]for every \(x\) in that set.
The value \(f(c)\) is the least function value under consideration.
Closed-Interval Extrema
On a closed interval
\[ [a,b], \]absolute extrema may occur at interior critical numbers or at endpoints.
A practical comparison procedure is:
- Find critical numbers inside \((a,b)\).
- Evaluate \(f\) at each critical number.
- Evaluate \(f(a)\) and \(f(b)\).
- Compare all resulting values.
Example: Absolute Extrema on a Closed Interval
Let
\[ f(x)=x^2-4x+5 \]on
\[ [0,5]. \]Differentiate:
\[ f'(x)=2x-4. \]The interior critical number is
\[ x=2. \]Evaluate:
\[ f(0)=5, \qquad f(2)=1, \qquad f(5)=10. \]Therefore, the absolute minimum is
\[ \boxed{1\text{ at }x=2} \]and the absolute maximum is
\[ \boxed{10\text{ at }x=5}. \]Why Endpoints Matter
A derivative test focuses mainly on interior behavior.
But on a restricted interval, an endpoint can contain the largest or smallest value even though there is no two-sided neighborhood around it.
This becomes especially important in optimization problems with practical constraints.
What Is Concavity?
Concavity describes how the slopes of a curve are changing.
A graph is concave up where tangent slopes are increasing.
A graph is concave down where tangent slopes are decreasing.
The second derivative provides a convenient test for this behavior.
Concave Up
If
\[ f''(x)>0 \]throughout an interval, then the derivative \(f'(x)\) is increasing there.
The original graph is concave up.
\[ \boxed{ f''(x)>0 \Rightarrow \text{concave up} }. \]Concave Down
If
\[ f''(x)<0 \]throughout an interval, then \(f'(x)\) is decreasing there.
The original graph is concave down.
\[ \boxed{ f''(x)<0 \Rightarrow \text{concave down} }. \]Concavity Is About Changing Slopes
Concavity should not be confused with whether a function is increasing or decreasing.
A function can be:
- increasing and concave up;
- increasing and concave down;
- decreasing and concave up;
- decreasing and concave down.
The first derivative describes the sign of the slope. The second derivative describes how that slope is changing.
Example: Concavity of a Parabola
Let
\[ f(x)=x^2. \]Then
\[ f'(x)=2x \]and
\[ f''(x)=2. \]Because
\[ 2>0 \]for every \(x\), the parabola is
\[ \boxed{\text{concave up everywhere}}. \]Example: Concave Down Parabola
Let
\[ f(x)=-x^2+4x. \]Then
\[ f'(x)=-2x+4 \]and
\[ f''(x)=-2. \]Since
\[ -2<0, \]the graph is
\[ \boxed{\text{concave down everywhere}}. \]Possible Inflection Points
An inflection point is a point where the graph changes concavity.
A common candidate occurs where
\[ f''(x)=0 \]or where \(f''(x)\) does not exist.
But merely solving \(f''(x)=0\) is not enough. Concavity must actually change across the point.
Example: Inflection Point of \(x^3\)
Let
\[ f(x)=x^3. \]Then
\[ f'(x)=3x^2 \]and
\[ f''(x)=6x. \]Set the second derivative equal to zero:
\[ 6x=0. \]Thus,
\[ x=0. \]For \(x<0\),
\[ f''(x)<0, \]while for \(x>0\),
\[ f''(x)>0. \]Concavity changes, so
\[ \boxed{(0,0)} \]is an inflection point.
Second-Derivative Sign Charts
We can build a sign chart for \(f''(x)\) just as we did for \(f'(x)\).
| Interval | Sign of \(f''(x)\) | Concavity |
|---|---|---|
| \((-\infty,c)\) | Negative | Concave down |
| \((c,\infty)\) | Positive | Concave up |
If the function is defined at \(x=c\), this sign change identifies an inflection point at
\[ (c,f(c)). \]Example: Concavity of a Cubic
Let
\[ f(x)=x^3-3x^2+2. \]Then
\[ f'(x)=3x^2-6x \]and
\[ f''(x)=6x-6. \]Solve
\[ 6x-6=0. \]This gives
\[ x=1. \]For \(x<1\), \(f''(x)<0\), so the graph is concave down.
For \(x>1\), \(f''(x)>0\), so the graph is concave up.
Finding the Inflection Point
Continue with
\[ f(x)=x^3-3x^2+2. \]The concavity changes at \(x=1\).
Evaluate:
\[ f(1)=1-3+2=0. \]Therefore, the inflection point is
\[ \boxed{(1,0)}. \]A Zero Second Derivative Need Not Be an Inflection Point
Consider
\[ f(x)=x^4. \]Then
\[ f'(x)=4x^3 \]and
\[ f''(x)=12x^2. \]At \(x=0\),
\[ f''(0)=0. \]But
\[ f''(x)>0 \]on both sides of \(0\).
The graph remains concave up, so the origin is not an inflection point.
The Second Derivative Test for Local Extrema
If
\[ f'(c)=0, \]the second derivative may help classify the stationary point.
- If \(f''(c)>0\), the graph is locally concave up and \(f(c)\) is a local minimum.
- If \(f''(c)<0\), the graph is locally concave down and \(f(c)\) is a local maximum.
- If \(f''(c)=0\), the test is inconclusive.
Why the Second Derivative Test Works
At a stationary point,
\[ f'(c)=0. \]If the graph is concave up there, the horizontal tangent sits near the bottom of a local bowl-like shape.
If the graph is concave down there, the horizontal tangent sits near the top of a local cap-like shape.
This geometric interpretation explains the second derivative test.
Example: Second Derivative Test
Let
\[ f(x)=x^2-6x+5. \]Then
\[ f'(x)=2x-6. \]The critical number is
\[ x=3. \]The second derivative is
\[ f''(x)=2. \]Since
\[ f''(3)=2>0, \]the critical point is a local minimum.
Second Derivative Test: Inconclusive Case
For
\[ f(x)=x^4, \]we have
\[ f'(x)=4x^3, \]so \(x=0\) is critical.
Also,
\[ f''(x)=12x^2 \]and
\[ f''(0)=0. \]The second derivative test gives no conclusion.
But first-derivative analysis shows \(f'\) changes from negative to positive, so \(x=0\) is a local minimum.
Increasing and Concave Up
Suppose
\[ f'(x)>0 \]and
\[ f''(x)>0. \]The function is increasing, and its positive slopes are becoming larger.
The graph rises and becomes steeper.
Increasing and Concave Down
Suppose
\[ f'(x)>0 \]but
\[ f''(x)<0. \]The function is increasing, but the positive slopes are becoming smaller.
The graph rises while flattening.
Decreasing and Concave Up
Suppose
\[ f'(x)<0 \]and
\[ f''(x)>0. \]The function is decreasing, but its slopes are becoming less negative.
The graph falls while flattening.
Decreasing and Concave Down
Suppose
\[ f'(x)<0 \]and
\[ f''(x)<0. \]The function is decreasing, and its slopes are becoming more negative.
The graph falls and becomes steeper.
Reading Shape from \(f'\) and \(f''\)
| Condition | Behavior |
|---|---|
| \(f'>0\) | Increasing |
| \(f'<0\) | Decreasing |
| \(f''>0\) | Concave up |
| \(f''<0\) | Concave down |
| \(f'=0\) | Horizontal tangent candidate |
| Concavity changes | Inflection-point behavior |
Example: Full Curve Analysis
Let
\[ f(x)=x^3-3x^2. \]First derivative:
\[ f'(x)=3x^2-6x = 3x(x-2). \]Critical numbers:
\[ x=0,\ 2. \]Second derivative:
\[ f''(x)=6x-6. \]Possible inflection input:
\[ x=1. \]Increasing and Decreasing Intervals in the Full Example
Use
\[ f'(x)=3x(x-2). \]| Interval | Sign of \(f'\) | Behavior |
|---|---|---|
| \((-\infty,0)\) | Positive | Increasing |
| \((0,2)\) | Negative | Decreasing |
| \((2,\infty)\) | Positive | Increasing |
Therefore, \(x=0\) is a local maximum input and \(x=2\) is a local minimum input.
Extrema in the Full Example
Evaluate the original function:
\[ f(0)=0, \] \[ f(2)=8-12=-4. \]Therefore, the local maximum point is
\[ \boxed{(0,0)} \]and the local minimum point is
\[ \boxed{(2,-4)}. \]Concavity in the Full Example
Use
\[ f''(x)=6x-6. \]For \(x<1\),
\[ f''(x)<0, \]so the graph is concave down.
For \(x>1\),
\[ f''(x)>0, \]so the graph is concave up.
Inflection Point in the Full Example
Concavity changes at
\[ x=1. \]Evaluate:
\[ f(1)=1-3=-2. \]Therefore, the inflection point is
\[ \boxed{(1,-2)}. \]First Derivative and Second Derivative Tell Different Stories
The first derivative answers:
- Is the curve rising or falling?
- Where is the tangent horizontal?
- Where can local extrema occur?
The second derivative answers:
- Are the slopes increasing or decreasing?
- Is the graph concave up or concave down?
- Where might concavity change?
Common Error: Using \(f(x)\) Instead of \(f'(x)\) for Increasing Behavior
The sign of
\[ f(x) \]tells whether the graph lies above or below the \(x\)-axis.
It does not directly tell whether the graph is increasing or decreasing.
For that, examine
\[ \boxed{f'(x)}. \]Common Error: Assuming \(f'(c)=0\) Guarantees an Extremum
A zero derivative gives a stationary point, but not necessarily a maximum or minimum.
Always check the sign of \(f'\) around the critical number or use another valid classification method.
Common Error: Calling \(f''(c)=0\) an Inflection Point Automatically
Solving
\[ f''(x)=0 \]finds candidates.
An inflection point requires an actual change in concavity.
Check the sign of \(f''\) on both sides of the candidate.
Common Error: Confusing Concavity with Increasing and Decreasing
A graph can be decreasing and concave up at the same time.
For example, a negative slope that moves from \(-5\) to \(-2\) is still negative, so the function is decreasing, but the slope is increasing, so the graph is concave up.
First- and second-derivative signs describe different layers of behavior.
Common Error: Ignoring Endpoints in Absolute-Extrema Problems
On a closed interval, the largest or smallest value may occur at an endpoint.
Checking only interior critical numbers can therefore miss an absolute extremum.
Common Error: Reporting Inputs When Points Are Requested
If \(x=2\) is where a local minimum occurs and
\[ f(2)=-3, \]then the minimum point is
\[ \boxed{(2,-3)}. \]The input \(2\) alone is the extremum location, not the full graph point.
Common Error: Testing a Critical Number Instead of an Interval
To determine the sign of \(f'\) between critical numbers, choose a test value inside each open interval.
Testing the critical number itself often gives
\[ f'(c)=0, \]which does not reveal the sign on either side.
A Reliable Procedure for Increasing and Decreasing Intervals
- Find \(f'(x)\).
- Find critical numbers.
- Use them to divide the domain into intervals.
- Choose one test value in each interval.
- Determine the sign of \(f'(x)\).
- Report intervals with \(f'>0\) as increasing.
- Report intervals with \(f'<0\) as decreasing.
A Reliable Procedure for Local Extrema
- Find the critical numbers.
- Build a sign chart for \(f'(x)\).
- Look for \(+\to-\) to identify local maxima.
- Look for \(-\to+\) to identify local minima.
- If there is no sign change, do not classify the point as a local extremum from the first derivative test.
- Evaluate \(f(c)\) if the extreme point or value is requested.
A Reliable Procedure for Concavity
- Find \(f''(x)\).
- Solve \(f''(x)=0\) and identify points where \(f''\) is undefined.
- Use those candidates to divide the domain into intervals.
- Test the sign of \(f''\) on each interval.
- Report \(f''>0\) intervals as concave up.
- Report \(f''<0\) intervals as concave down.
- Identify inflection points only where concavity actually changes and the original function is defined.
A Reliable Procedure for Absolute Extrema on a Closed Interval
- Find critical numbers in the interior of the interval.
- Evaluate the original function at each interior critical number.
- Evaluate the original function at both endpoints.
- Compare all values.
- The greatest value is the absolute maximum.
- The least value is the absolute minimum.
Practice 1: Increasing and Decreasing
Let
\[ f(x)=x^2-4x. \]Find the intervals where \(f\) is increasing and decreasing.
Practice 2: Local Extremum
For
\[ f(x)=x^2-4x, \]identify the local extremum and its point.
Practice 3: Two Critical Numbers
Let
\[ f(x)=x^3-12x. \]Determine the increasing and decreasing intervals.
Practice 4: Classify the Critical Points
For
\[ f(x)=x^3-12x, \]classify each critical point as a local maximum or local minimum.
Practice 5: Concavity
Let
\[ f(x)=x^3+3x^2. \]Find the intervals where the graph is concave up and concave down.
Practice 6: Inflection Point
For
\[ f(x)=x^3+3x^2, \]find the inflection point.
Practice 7: Second Derivative Test
Let
\[ f(x)=2x^2-8x+1. \]Use the second derivative test to classify its critical point.
Practice 8: Closed-Interval Absolute Extrema
Let
\[ f(x)=x^2-2x+3 \]on
\[ [0,4]. \]Find the absolute maximum and absolute minimum values.
Practice 9: No Inflection Point
Explain why
\[ f(x)=x^4 \]has no inflection point at \(x=0\), even though \(f''(0)=0\).
Practice 10: Identify the Error
A student finds
\[ f''(2)=0 \]and immediately concludes that \(x=2\) is an inflection point.
Explain what additional information is required.
Practice Answers
Practice 1
\[ f'(x)=2x-4. \]The critical number is
\[ x=2. \]For \(x<2\),
\[ f'(x)<0, \]so \(f\) is decreasing on
\[ \boxed{(-\infty,2)}. \]For \(x>2\),
\[ f'(x)>0, \]so \(f\) is increasing on
\[ \boxed{(2,\infty)}. \]Practice 2
The derivative changes from negative to positive at \(x=2\), so there is a local minimum.
\[ f(2)=4-8=-4. \]The local minimum point is
\[ \boxed{(2,-4)}. \]Practice 3
\[ f'(x)=3x^2-12 = 3(x-2)(x+2). \]The critical numbers are
\[ -2,\ 2. \]The sign pattern is
\[ +,\ -,\ +. \]Therefore, \(f\) is increasing on
\[ \boxed{(-\infty,-2)\cup(2,\infty)} \]and decreasing on
\[ \boxed{(-2,2)}. \]Practice 4
At \(x=-2\), the derivative changes from positive to negative, so there is a local maximum.
\[ f(-2)=-8+24=16. \]At \(x=2\), the derivative changes from negative to positive, so there is a local minimum.
\[ f(2)=8-24=-16. \]Thus, the local maximum point is
\[ \boxed{(-2,16)} \]and the local minimum point is
\[ \boxed{(2,-16)}. \]Practice 5
\[ f'(x)=3x^2+6x, \] \[ f''(x)=6x+6. \]Set
\[ 6x+6=0, \]giving
\[ x=-1. \]For \(x<-1\), \(f''(x)<0\), so the graph is concave down.
For \(x>-1\), \(f''(x)>0\), so the graph is concave up.
\[ \boxed{ \text{Concave down on }(-\infty,-1), \quad \text{concave up on }(-1,\infty) }. \]Practice 6
Concavity changes at \(x=-1\).
\[ f(-1)=-1+3=2. \]Therefore, the inflection point is
\[ \boxed{(-1,2)}. \]Practice 7
\[ f'(x)=4x-8. \]The critical number is
\[ x=2. \]The second derivative is
\[ f''(x)=4. \]Since
\[ f''(2)=4>0, \]the critical point is a local minimum.
Practice 8
\[ f'(x)=2x-2. \]The interior critical number is
\[ x=1. \]Evaluate:
\[ f(0)=3, \qquad f(1)=2, \qquad f(4)=11. \]Therefore, the absolute minimum value is
\[ \boxed{2} \]at \(x=1\), and the absolute maximum value is
\[ \boxed{11} \]at \(x=4\).
Practice 9
\[ f''(x)=12x^2. \]Although
\[ f''(0)=0, \]the second derivative is positive on both sides of \(0\). The graph is concave up on both sides, so concavity does not change.
Therefore, \(x=0\) is not an inflection point.
Practice 10
The equation
\[ f''(2)=0 \]identifies only a possible inflection input.
The student must check the sign of \(f''(x)\) on both sides of \(2\) and verify that the original function is defined there.
An inflection point requires an actual change in concavity.
What to Remember
First-derivative signs describe direction:
\[ \boxed{ f'(x)>0 \Rightarrow \text{increasing} } \] \[ \boxed{ f'(x)<0 \Rightarrow \text{decreasing} }. \]First-derivative sign changes classify many critical points:
\[ \boxed{ +\to- \Rightarrow \text{local maximum} } \] \[ \boxed{ -\to+ \Rightarrow \text{local minimum} }. \]Second-derivative signs describe concavity:
\[ \boxed{ f''(x)>0 \Rightarrow \text{concave up} } \] \[ \boxed{ f''(x)<0 \Rightarrow \text{concave down} }. \]- critical numbers divide the domain into intervals for first-derivative sign analysis;
- local extrema depend on nearby behavior;
- absolute extrema require comparison across the full stated domain or interval;
- endpoints must be checked in closed-interval problems;
- inflection points require an actual change in concavity;
- \(f''(c)=0\) alone does not prove an inflection point;
- the second derivative test can classify some stationary points, but \(f''(c)=0\) makes that test inconclusive.
Where We Go Next
Derivative analysis lets us identify where a function rises, falls, reaches local highs or lows, and changes its bending behavior. These tools become especially useful when a practical problem asks us to choose the largest, smallest, fastest, cheapest, shortest, or most efficient possible value under stated conditions.
The next lesson, Introductory Optimization, turns extrema analysis into a problem-solving method for maximizing and minimizing quantities subject to practical constraints.