The Central Idea
The sign of the first derivative controls local direction:
\[ f'(x)>0 \quad\Longrightarrow\quad f\text{ is increasing locally}, \] \[ f'(x)<0 \quad\Longrightarrow\quad f\text{ is decreasing locally}. \]The sign of the second derivative controls local bending:
\[ f''(x)>0 \quad\Longrightarrow\quad f\text{ is concave up}, \] \[ f''(x)<0 \quad\Longrightarrow\quad f\text{ is concave down}. \]The main task is to locate where these signs can change and then test the intervals between those locations.
From Pointwise Slopes to Interval Behavior
A derivative value such as
\[ f'(2)=5 \]tells us the slope at one point.
But an inequality such as
\[ f'(x)>0 \qquad \text{for }1Differential analysis often shifts from computing derivatives to studying their signs.
What Does Increasing Mean?
A function is increasing on an interval when larger inputs correspond to larger outputs.
In derivative-based analysis, a common practical test is:
\[ \boxed{ f'(x)>0 \text{ throughout an interval} \Longrightarrow f\text{ is increasing there} }. \]Positive slope means the graph rises as we move from left to right.
What Does Decreasing Mean?
A function is decreasing on an interval when larger inputs correspond to smaller outputs.
A standard derivative test is:
\[ \boxed{ f'(x)<0 \text{ throughout an interval} \Longrightarrow f\text{ is decreasing there} }. \]Negative slope means the graph falls as we move from left to right.
Monotonicity
Monotonicity describes whether a function consistently moves in one direction over an interval.
In this lesson, the main monotonicity categories are:
- increasing;
- decreasing.
Derivative sign analysis provides a systematic way to identify these intervals.
What Is a Critical Number?
A number \(c\) in the domain of \(f\) is a critical number if
\[ f'(c)=0 \]or
\[ f'(c) \]does not exist.
The requirement that \(c\) lie in the domain of \(f\) matters.
Critical Number versus Critical Point
The input value
\[ c \]is a critical number.
The corresponding point on the graph,
\[ (c,f(c)), \]is often called a critical point.
Keeping the input value and graph point distinct makes later classification easier.
Why Critical Numbers Matter
Critical numbers are locations where the usual slope behavior may change.
They can occur at:
- horizontal tangents;
- corners;
- cusps;
- vertical tangents;
- other points where the derivative fails to exist.
They are natural candidates for changes in monotonicity or local extreme behavior.
A Critical Number Need Not Be an Extremum
Consider
\[ f(x)=x^3. \]Then
\[ f'(x)=3x^2. \]Since
\[ f'(0)=0, \]\(x=0\) is a critical number.
But \(x^3\) keeps increasing through \(0\), so the critical point is not a local maximum or local minimum.
Critical Numbers from \(f'(x)=0\)
Suppose
\[ f(x)=x^3-3x. \]Differentiate:
\[ f'(x)=3x^2-3. \]Set the derivative equal to zero:
\[ 3x^2-3=0. \]Factor:
\[ 3(x-1)(x+1)=0. \]Therefore, the critical numbers are
\[ \boxed{x=-1,\quad x=1}. \]Critical Numbers Where the Derivative Is Undefined
Consider
\[ f(x)=|x|. \]The function is defined at \(x=0\), but the derivative does not exist there because the left-hand and right-hand slopes differ.
Therefore,
\[ \boxed{x=0} \]is a critical number.
Domain Breaks Are Not Automatically Critical Numbers
Consider
\[ f(x)=\frac1x. \]The derivative is
\[ f'(x)=-\frac1{x^2}. \]The derivative is undefined at \(x=0\), but so is \(f\).
Since \(0\) is not in the domain of \(f\), it is not a critical number of this function.
It is still an important boundary when we divide the domain into intervals.
Partitioning the Domain
Critical numbers and domain breaks divide the domain into intervals on which the sign of the derivative can often be determined consistently.
For example, if critical numbers occur at
\[ x=-1 \qquad\text{and}\qquad x=1, \]then the real line is divided into
\[ (-\infty,-1), \qquad (-1,1), \qquad (1,\infty). \]We test \(f'(x)\) on each interval.
The First-Derivative Sign Chart
A sign chart organizes the sign of \(f'(x)\) over intervals.
Suppose
\[ f'(x)=3(x-1)(x+1). \]The sign pattern is
\[ \begin{array}{c|ccccc} x & (-\infty,-1) & -1 & (-1,1) & 1 & (1,\infty)\\ \hline f'(x) & + & 0 & - & 0 & + \end{array} \]Therefore, \(f\) increases, then decreases, then increases.
Example: Monotonicity of \(x^3-3x\)
Let
\[ f(x)=x^3-3x. \]We found
\[ f'(x)=3(x-1)(x+1). \]Test one point in each interval.
For \(x=-2\),
\[ f'(-2)>0. \]For \(x=0\),
\[ f'(0)<0. \]For \(x=2\),
\[ f'(2)>0. \]Monotonicity Intervals
Therefore,
\[ \boxed{ f\text{ is increasing on } (-\infty,-1)\cup(1,\infty) } \]and
\[ \boxed{ f\text{ is decreasing on } (-1,1) }. \]The derivative sign tells the directional behavior directly.
The First-Derivative Test
A critical number can be classified by checking how the sign of \(f'\) changes across it.
- If \(f'\) changes from positive to negative, the function changes from increasing to decreasing.
- If \(f'\) changes from negative to positive, the function changes from decreasing to increasing.
- If the sign does not change, there is no turning behavior of that kind.
This is the first-derivative test.
Positive to Negative
If
\[ f'(x)>0 \]immediately to the left of \(c\), and
\[ f'(x)<0 \]immediately to the right, then the graph rises into \(c\) and falls away from \(c\).
This indicates a local high point.
Negative to Positive
If
\[ f'(x)<0 \]immediately to the left of \(c\), and
\[ f'(x)>0 \]immediately to the right, then the graph falls into \(c\) and rises away from \(c\).
This indicates a local low point.
No Sign Change
If \(f'(x)\) has the same sign on both sides of a critical number, the function continues moving in the same direction.
For example, for
\[ f(x)=x^3, \]we have
\[ f'(x)=3x^2>0 \]for all \(x\neq0\).
So the function increases on both sides of the critical number \(0\).
What Is Concavity?
Concavity describes how the slope changes as \(x\) increases.
A function is concave up on an interval when its slopes are increasing.
A function is concave down on an interval when its slopes are decreasing.
Second Derivative and Concavity
Since \(f''\) measures the rate of change of \(f'\),
\[ \boxed{ f''(x)>0 \Longrightarrow f\text{ is concave up} } \]and
\[ \boxed{ f''(x)<0 \Longrightarrow f\text{ is concave down} }. \]The second derivative therefore converts changing-slope information into a sign test.
Concave Up Does Not Mean Increasing
Concavity and monotonicity describe different features.
A function can be decreasing and concave up at the same time.
For example,
\[ f(x)=e^{-x} \]satisfies
\[ f'(x)=-e^{-x}<0 \]and
\[ f''(x)=e^{-x}>0. \]So it is decreasing but concave up.
Concave Down Does Not Mean Decreasing
A function can also be increasing and concave down.
For example,
\[ f(x)=\sqrt{x} \qquad(x>0) \]satisfies
\[ f'(x)=\frac1{2\sqrt{x}}>0 \]and
\[ f''(x) = -\frac1{4x^{3/2}}<0. \]The function rises, but its slope becomes smaller.
What Is an Inflection Point?
An inflection point is a point on the graph where the concavity changes.
The essential condition is a genuine change such as
\[ \text{concave up} \longrightarrow \text{concave down} \]or
\[ \text{concave down} \longrightarrow \text{concave up}. \]A zero of \(f''\) is only a candidate until the sign change is verified.
Finding Inflection Candidates
Possible concavity changes often occur where
\[ f''(x)=0 \]or where
\[ f''(x) \]is undefined.
These values divide the domain into intervals for a second-derivative sign chart.
Example: Concavity of \(x^3-3x\)
Let
\[ f(x)=x^3-3x. \]We have
\[ f'(x)=3x^2-3 \]and
\[ f''(x)=6x. \]Set the second derivative equal to zero:
\[ 6x=0. \]Thus,
\[ x=0 \]is an inflection candidate.
Second-Derivative Sign Chart
For
\[ f''(x)=6x, \]we have
\[ f''(x)<0 \qquad \text{when }x<0 \]and
\[ f''(x)>0 \qquad \text{when }x>0. \]Therefore,
\[ \boxed{ f\text{ is concave down on }(-\infty,0) } \]and
\[ \boxed{ f\text{ is concave up on }(0,\infty) }. \]Confirming the Inflection Point
The concavity changes at \(x=0\), so \(x=0\) corresponds to an inflection point.
Since
\[ f(0)=0, \]the inflection point is
\[ \boxed{(0,0)}. \]\(f''(c)=0\) Is Not Enough
Consider
\[ f(x)=x^4. \]Then
\[ f''(x)=12x^2. \]At \(x=0\),
\[ f''(0)=0. \]But
\[ f''(x)>0 \]on both sides of \(0\).
Concavity does not change, so \(x=0\) is not an inflection point.
An Undefined Second Derivative Can Still Mark an Inflection
A concavity change can occur where \(f''\) does not exist.
The decisive question is not whether \(f''(c)=0\), but whether the concavity actually changes across the point and the function has a point there to classify.
This parallels the first-derivative lesson: undefined derivatives can still signal meaningful geometric behavior.
First and Second Derivatives Answer Different Questions
| Derivative Information | Primary Question |
|---|---|
| \(f'(x)>0\) | Is the function increasing? |
| \(f'(x)<0\) | Is the function decreasing? |
| \(f''(x)>0\) | Is the function concave up? |
| \(f''(x)<0\) | Is the function concave down? |
Monotonicity describes direction. Concavity describes changing slope.
Four Common Local Behavior Combinations
A function can exhibit any of these combinations:
- increasing and concave up;
- increasing and concave down;
- decreasing and concave up;
- decreasing and concave down.
First- and second-derivative signs allow us to distinguish them.
Increasing and Concave Up
If
\[ f'(x)>0 \]and
\[ f''(x)>0, \]then the function is rising and its slopes are becoming more positive.
Visually, the graph rises and steepens.
Increasing and Concave Down
If
\[ f'(x)>0 \]and
\[ f''(x)<0, \]then the function is rising but its slopes are becoming smaller.
Visually, the graph rises while flattening.
Decreasing and Concave Up
If
\[ f'(x)<0 \]and
\[ f''(x)>0, \]then the function is falling, but its negative slopes are becoming less negative.
Visually, the graph falls while flattening.
Decreasing and Concave Down
If
\[ f'(x)<0 \]and
\[ f''(x)<0, \]then the function is falling and its slopes are becoming more negative.
Visually, the graph falls and steepens downward.
Example: Full Sign Analysis of a Quartic
Consider
\[ f(x)=x^4-4x^2. \]First derivative:
\[ f'(x)=4x^3-8x. \]Factor:
\[ f'(x)=4x(x^2-2). \]Therefore, the critical numbers are
\[ \boxed{ x=-\sqrt2,\quad 0,\quad \sqrt2 }. \]Monotonicity of the Quartic
Test the sign of
\[ f'(x)=4x(x^2-2) \]on the intervals determined by the critical numbers.
The sign pattern is
\[ -, \quad +,\quad -,\quad + \]across
\[ (-\infty,-\sqrt2),\quad (-\sqrt2,0),\quad (0,\sqrt2),\quad (\sqrt2,\infty). \]Thus,
\[ \boxed{ f\text{ decreases on } (-\infty,-\sqrt2)\cup(0,\sqrt2) } \]and
\[ \boxed{ f\text{ increases on } (-\sqrt2,0)\cup(\sqrt2,\infty) }. \]Concavity of the Quartic
Differentiate again:
\[ f''(x)=12x^2-8. \]Set
\[ 12x^2-8=0. \]Then
\[ 12x^2=8, \] \[ x^2=\frac23. \]Therefore, the concavity candidates are
\[ \boxed{ x=\pm\sqrt{\frac23} }. \]Concavity Intervals of the Quartic
Since
\[ f''(x)=12x^2-8, \]the second derivative is positive when
\[ |x|>\sqrt{\frac23} \]and negative when
\[ |x|<\sqrt{\frac23}. \]Therefore,
\[ \boxed{ f\text{ is concave up on } \left(-\infty,-\sqrt{\frac23}\right) \cup \left(\sqrt{\frac23},\infty\right) } \]and
\[ \boxed{ f\text{ is concave down on } \left(-\sqrt{\frac23},\sqrt{\frac23}\right) }. \]Inflection Points of the Quartic
Because the sign of \(f''\) changes at both candidates, both produce inflection points.
For
\[ x^2=\frac23, \]we have
\[ x^4=\frac49. \]Therefore,
\[ f(x) = \frac49-4\left(\frac23\right) = \frac49-\frac{24}{9} = -\frac{20}{9}. \]The inflection points are
\[ \boxed{ \left( -\sqrt{\frac23}, -\frac{20}{9} \right) \quad\text{and}\quad \left( \sqrt{\frac23}, -\frac{20}{9} \right) }. \]Sign Charts Are More Reliable Than Guessing from Formulas
A complicated derivative may not make its sign obvious at first glance.
The reliable process is:
- factor when possible;
- identify zeros and undefined points;
- partition the domain;
- choose a test point in each interval;
- record the sign;
- translate the sign into function behavior.
Why One Test Point per Interval Usually Works
Once all zeros and undefined points of the derivative have been used to divide the domain, the derivative cannot change sign inside one of the remaining intervals without passing through another zero or discontinuity.
Therefore, a single correctly chosen test point can determine the sign on the entire interval, assuming the derivative is continuous there.
Polynomial Derivatives Are Especially Convenient
For polynomial functions, derivatives are also polynomials.
Polynomial derivatives are continuous everywhere, so their signs can change only where they equal zero.
This makes sign-chart analysis especially systematic for polynomial functions.
Rational Functions Require Domain Awareness
For a rational function, both zeros and denominator restrictions may divide the domain.
Consider
\[ f(x)=\frac{x}{x-1}. \]The point
\[ x=1 \]is not in the domain, but it still splits the domain into separate intervals for analysis.
Example: Monotonicity with a Domain Break
For
\[ f(x)=\frac{x}{x-1}, \]the derivative is
\[ f'(x) = \frac{(x-1)-x}{(x-1)^2} = -\frac1{(x-1)^2}. \]Therefore,
\[ f'(x)<0 \]for every \(x\neq1\).
The function is decreasing on each domain interval:
\[ \boxed{ (-\infty,1) \qquad\text{and}\qquad (1,\infty) }. \]Do Not Combine Across a Domain Break
Even if the derivative has the same sign on both sides of a discontinuity, we should respect the disconnected domain.
For
\[ f(x)=\frac{x}{x-1}, \]we report decreasing behavior separately on
\[ (-\infty,1) \]and
\[ (1,\infty), \]rather than pretending the function forms one continuous decreasing branch across \(x=1\).
Derivative Information Can Be Given Without a Formula for \(f\)
Sometimes we know only the derivative.
Suppose
\[ f'(x)=(x-2)(x+5). \]Then the critical numbers are
\[ -5 \qquad\text{and}\qquad 2. \]The sign of \(f'\) is positive outside these roots and negative between them.
Therefore, we can determine monotonicity without ever finding a formula for \(f(x)\).
Interpreting a Graph of \(f'\)
If the graph of \(f'\) lies above the \(x\)-axis, then
\[ f'(x)>0, \]so \(f\) is increasing.
If the graph of \(f'\) lies below the \(x\)-axis, then
\[ f'(x)<0, \]so \(f\) is decreasing.
Crossings of the derivative graph through the axis can signal changes in monotonicity.
Interpreting the Slope of \(f'\)
Since
\[ f''(x) \]is the derivative of \(f'\), the slope of the graph of \(f'\) tells us about concavity of \(f\).
- If \(f'\) is increasing, then \(f''>0\), so \(f\) is concave up.
- If \(f'\) is decreasing, then \(f''<0\), so \(f\) is concave down.
Physical Interpretation
If \(s(t)\) is position, then
\[ s'(t)=v(t) \]is velocity and
\[ s''(t)=a(t) \]is acceleration.
Thus:
- \(s'(t)>0\) means position is increasing;
- \(s'(t)<0\) means position is decreasing;
- \(s''(t)>0\) means velocity is increasing;
- \(s''(t)<0\) means velocity is decreasing.
Monotonicity and Accumulation
If a quantity has positive derivative over an interval, it accumulates upward over that interval.
If its derivative is negative, it moves downward.
This general principle applies to position, population, revenue, energy, concentration, temperature, and many other modeled quantities.
Concavity as Changing Marginal Effect
In applications, \(f'(x)\) often represents a marginal effect.
Then \(f''(x)\) tells us whether that marginal effect is increasing or decreasing.
For example:
- \(f''>0\) can represent accelerating growth;
- \(f''<0\) can represent diminishing marginal gains;
- a concavity change can mark a shift between those regimes.
A Reliable Procedure for Monotonicity
- Determine the domain of \(f\).
- Compute \(f'(x)\).
- Find where \(f'(x)=0\).
- Find where \(f'(x)\) is undefined.
- Keep only domain points as critical numbers.
- Use critical numbers and domain breaks to partition the domain.
- Test the sign of \(f'\) on each interval.
- Report increasing intervals where \(f'>0\).
- Report decreasing intervals where \(f'<0\).
A Reliable Procedure for Concavity
- Compute \(f''(x)\).
- Find where \(f''(x)=0\).
- Find where \(f''(x)\) is undefined.
- Use these values and domain breaks to partition the domain.
- Test the sign of \(f''\) on each interval.
- Report concave-up intervals where \(f''>0\).
- Report concave-down intervals where \(f''<0\).
- Confirm inflection points only where concavity actually changes.
Common Error: Solving Only \(f'(x)=0\)
Critical numbers can also occur where
\[ f'(x) \]does not exist while \(f(x)\) remains defined.
Corners, cusps, and vertical tangents can therefore be missed if we search only for zeros of the derivative.
Common Error: Calling a Domain Break a Critical Number
If \(f(c)\) is undefined, then \(c\) is not a critical number of \(f\).
The value may still be important for splitting the domain, but the definition of critical number requires
\[ c\in\operatorname{Dom}(f). \]Common Error: Assuming Every Critical Point Is a Maximum or Minimum
The equation
\[ f'(c)=0 \]identifies a candidate, not a classification.
The derivative sign must be examined around \(c\) to determine whether the function changes direction.
Common Error: Confusing Increasing with Concave Up
Increasing means
\[ f'(x)>0. \]Concave up means
\[ f''(x)>0. \]These are separate statements.
Common Error: Declaring an Inflection Point from \(f''(c)=0\)
A zero of the second derivative is only a candidate.
The concavity must actually change across the point.
For
\[ f(x)=x^4, \]\(f''(0)=0\), but there is no concavity change.
Common Error: Forgetting to Report Intervals
Monotonicity and concavity describe interval behavior.
Therefore, answers should usually be reported with interval notation such as
\[ (-\infty,-1) \qquad\text{or}\qquad (2,5). \]A list of isolated critical numbers is not a substitute for interval analysis.
Practice 1: Critical Numbers
Find the critical numbers of
\[ f(x)=x^3-12x. \]Practice 2: Monotonicity
Determine the intervals where
\[ f(x)=x^3-12x \]is increasing and decreasing.
Practice 3: First-Derivative Sign Change
Suppose \(f'(x)\) changes from negative to positive at \(x=4\). Describe the local behavior of \(f\) near \(x=4\).
Practice 4: Concavity
Find the intervals of concavity for
\[ f(x)=x^3+6x^2. \]Practice 5: Inflection Point
Find the inflection point of
\[ f(x)=x^3+6x^2. \]Practice 6: No Inflection
Explain why \(x=0\) is not an inflection point of
\[ f(x)=x^4. \]Practice 7: Domain Break
For
\[ f(x)=\frac1{x-2}, \]is \(x=2\) a critical number? Explain.
Practice 8: Derivative Given Directly
Suppose
\[ f'(x)=(x+2)(x-3). \]Determine where \(f\) is increasing and decreasing.
Practice 9: Second Derivative Given Directly
Suppose
\[ f''(x)=x(x-5). \]Determine where \(f\) is concave up and concave down.
Practice 10: Identify the Error
A student solves
\[ f''(x)=0 \]and automatically labels every solution an inflection point.
Explain what additional step is required.
Practice Answers
Practice 1
Differentiate:
\[ f'(x)=3x^2-12. \]Set equal to zero:
\[ 3x^2-12=0. \] \[ x^2=4. \]Therefore,
\[ \boxed{x=-2,\quad x=2}. \]Practice 2
Factor:
\[ f'(x)=3(x-2)(x+2). \]The sign is positive on
\[ (-\infty,-2) \]and
\[ (2,\infty), \]and negative on
\[ (-2,2). \]Therefore,
\[ \boxed{ f\text{ increases on } (-\infty,-2)\cup(2,\infty) } \]and
\[ \boxed{ f\text{ decreases on } (-2,2) }. \]Practice 3
Negative derivative means the function decreases before \(x=4\), while positive derivative means it increases after \(x=4\).
Therefore, the function changes from decreasing to increasing and has a local low point there.
Practice 4
Differentiate twice:
\[ f'(x)=3x^2+12x, \] \[ f''(x)=6x+12. \]Set
\[ 6x+12=0. \]Then
\[ x=-2. \]For \(x<-2\),
\[ f''(x)<0, \]while for \(x>-2\),
\[ f''(x)>0. \]Therefore,
\[ \boxed{ f\text{ is concave down on }(-\infty,-2) } \]and
\[ \boxed{ f\text{ is concave up on }(-2,\infty) }. \]Practice 5
The concavity changes at \(x=-2\).
Evaluate:
\[ f(-2)=(-2)^3+6(-2)^2=-8+24=16. \]Therefore, the inflection point is
\[ \boxed{(-2,16)}. \]Practice 6
For
\[ f(x)=x^4, \]we have
\[ f''(x)=12x^2. \]Although
\[ f''(0)=0, \]the second derivative is positive on both sides of \(0\).
The function is concave up on both sides, so there is no inflection.
Practice 7
No.
The function
\[ f(x)=\frac1{x-2} \]is undefined at \(x=2\).
Since a critical number must belong to the domain of \(f\), \(2\) is not a critical number. It is a domain break that still divides the intervals of analysis.
Practice 8
The zeros of
\[ f'(x)=(x+2)(x-3) \]are
\[ x=-2 \qquad\text{and}\qquad x=3. \]The derivative is positive outside the roots and negative between them.
Therefore,
\[ \boxed{ f\text{ increases on } (-\infty,-2)\cup(3,\infty) } \]and
\[ \boxed{ f\text{ decreases on } (-2,3) }. \]Practice 9
The zeros of
\[ f''(x)=x(x-5) \]are
\[ 0 \qquad\text{and}\qquad 5. \]The second derivative is positive on
\[ (-\infty,0) \]and
\[ (5,\infty), \]and negative on
\[ (0,5). \]Therefore,
\[ \boxed{ f\text{ is concave up on } (-\infty,0)\cup(5,\infty) } \]and
\[ \boxed{ f\text{ is concave down on } (0,5) }. \]Practice 10
Solving
\[ f''(x)=0 \]identifies only candidates.
The student must test the sign of \(f''\) on both sides and verify that the concavity actually changes.
What to Remember
Critical numbers occur at domain points where
\[ \boxed{ f'(c)=0 \quad\text{or}\quad f'(c)\text{ does not exist} }. \]First-derivative signs determine monotonicity:
\[ \boxed{ f'(x)>0 \Longrightarrow f\text{ increasing} } \] \[ \boxed{ f'(x)<0 \Longrightarrow f\text{ decreasing} }. \]Second-derivative signs determine concavity:
\[ \boxed{ f''(x)>0 \Longrightarrow f\text{ concave up} } \] \[ \boxed{ f''(x)<0 \Longrightarrow f\text{ concave down} }. \]Inflection points require an actual change in concavity.
The most reliable workflow is:
- find the relevant zeros and undefined points;
- partition the domain;
- build derivative sign charts;
- translate signs into qualitative behavior.
Where We Go Next
Critical-point, monotonicity, and concavity analysis gives us the raw derivative information needed to understand the shape of a function.
The next lesson, Extrema and Curve Sketching, assembles that information with intercepts, endpoint behavior, local and absolute extrema, asymptotic behavior, and other structural features to build a coherent qualitative graph.