Graph Analysis with Derivatives: Extrema & Curves

The Derivative at a Point

The value

f'(a)

is a number—the slope of the tangent line to the graph of

f

at the point

(a, f(a))

. A positive value means the function is increasing at that instant. A negative value means it is decreasing. A value of zero means the graph is momentarily flat.

The magnitude

|f'(a)|

measures steepness. A derivative of

10

at a point means the function is climbing steeply; a derivative of

0.01

means it is nearly flat. The sign gives direction; the size gives intensity.

At points where

f'(a)

does not exist—corners, cusps, vertical tangents, or discontinuities—the function has no well-defined instantaneous rate of change. These points require separate treatment under differentiability.

Tangent Lines

The tangent line to

f

at

x = a

passes through

(a, f(a))

with slope

f'(a)

. Its equation in point-slope form is:

Tangent Line Equation

y - f(a) = f'(a)(x - a) \qquad y = f(a) + f'(a)(x - a)

Learn more about this formula: Tangent Line Equation →

Pitfall: Leaving x in the slope

Putting the derivative function into the formula for

f(x) = x^2

at

a = 3

gives

y - 9 = 2x(x - 3)

, a parabola, not a line. The slope is the number

f'(3) = 6

:

y - 9 = 6(x - 3)

, so

y = 6x - 9

.

This line is the best linear approximation to

f

near

x = a

. For values of

x

close to

a

, the tangent line and the curve are nearly indistinguishable. This is the geometric foundation of linear approximation.

The normal line at the same point is perpendicular to the tangent. If

f'(a) \neq 0

, its slope is

-1/f'(a)

. If

f'(a) = 0

, the tangent is horizontal and the normal is vertical.

Normal Line Equation

y - f(a) = -\frac{1}{f'(a)}(x - a)

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Tangent at P with slope f′(c) = 1.89

The line touches the curve at P and nowhere else nearby, and its slope 1.89 is the derivative at that point; the point-slope equation of the section is this line written out. Where the curve is steeper the line tilts more, where the curve flattens the line goes level. Drag P along the curve and read the slope on the tangent line at a point tool.

The tangent line is the local stand-in for the curve, and its slope is the derivative made visible.

Notation of Graph Analysis

Notation

Notation of Graph Analysis

Three working conventions the tests below lean on: the reserved letter

c

, the sign-change arrow, and inequalities written on whole functions.

f'

and

f'(a)

— the derivative function;

f''

— higher-order derivatives; the

\implies

arrow — logical implication.

c

The critical number c

A witness letter: a specific but unnamed point.

x

ranges,

a

is a chosen anchor,

c

is found — by solving

f'(c) = 0

, by spotting where

f'

fails to exist (Critical Points below), or handed over by the Mean Value Theorem.

Cases

f'(c) = 0

and “

f'(c)

does not exist” are both critical — forgetting the second loses corner extrema like

|x|

at

0

.

Also written

x_0

— the subscript-zero convention, standard in Russian and European texts for the same distinguished point.

Do not confuseThe capital

+C

of the indefinite integral — an arbitrary constant, free to be anything; this

c

is the opposite, one fixed point.

f': \; + \to -

f prime changes from positive to negative

Sign immediately left of

c

, then immediately right. The arrow is spatial — crossing

c

— not the limit arrow of two-sided limits, which shares the glyph.

Cases

+ \to -

— local max;

- \to +

— local min; no change — neither, as in The First Derivative Test below.

Also writtenSign charts: intervals in a row, signs beneath. French texts draw the tableau de variations, with ↗ and ↘ arrows carrying the same information pictorially.

Do not confuseA statement about

f'(c)

itself. The test reads the sign around

c

, never at it —

f'(c)

may be zero or fail to exist and the test still runs.

f' > 0

on

(a,b)

f prime is positive on the interval

An inequality on a whole function:

f'(x) > 0

for every

x

in

(a,b)

, argument suppressed. The quantifier is hidden in the word “on”.

CasesAt a single point —

f''(c) > 0

, the second-derivative test's hypothesis. On an interval — monotonicity and concavity claims. The two read alike and assert very different amounts.

Also writtenSpelled out:

\forall x \in (a,b),\; f'(x) > 0

— the universal quantifier made visible.

Do not confuse“

f'

is positive somewhere.” The suppressed quantifier is every, not some — one sign check per interval only suffices after the zeros of

f'

have been found, which is what makes critical points the required first step.

All calculus symbols →↑ Derivatives

Increasing and Decreasing Functions

The sign of

f'

on an interval determines the direction of

f

on that interval.

If

f'(x) > 0

for all

x

in an open interval

(a, b)

, then

f

is strictly increasing on

(a, b)

: larger inputs produce larger outputs. If

f'(x) < 0

on

(a, b)

, then

f

is strictly decreasing: larger inputs produce smaller outputs.

Sign of First Derivative

f'(x) > 0 \implies f \text{ increasing}, \quad f'(x) < 0 \implies f \text{ decreasing}

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The proof relies on the Mean Value Theorem. For any two points

x_1 < x_2

in

(a, b)

, there exists

c

between them with

f(x_2) - f(x_1) = f'(c)(x_2 - x_1)

. If

f'(c) > 0

and

x_2 - x_1 > 0

, then

f(x_2) - f(x_1) > 0

, confirming

f

is increasing.

To find where a function increases or decreases: solve

f'(x) = 0

and identify where

f'

is undefined, then test the sign of

f'

in each resulting interval.

Pitfall: Joining intervals across a break

For

f(x) = -\frac{1}{x}

,

f'(x) = \frac{1}{x^2} > 0

wherever it exists, yet "increasing on

(-\infty, 0) \cup (0, \infty)

" is false:

f(-1) = 1 > f(1) = -1

. The function is increasing on

(-\infty, 0)

and on

(0, \infty)

, separately.

The points found this way split the domain into open intervals. On each interval

f'

keeps a single sign, so one test value inside it is enough: for

f(x) = x^3 - 3x

, the points

x = \pm 1

give three intervals, and

f'(-2) = 9

,

f'(0) = -3

and

f'(2) = 9

show that

f

increases, decreases, then increases again. A value of

x

where

f

itself is undefined also splits the domain, since

f'

cannot exist there either.

On each interval where

f'

keeps one sign, that sign alone tells whether

f

rises or falls.

Critical Points

A critical point of

f

is a value

x = c

in the domain of

f

where either

f'(c) = 0

or

f'(c)

does not exist.

Critical Point Condition

f'(c) = 0 \quad \text{or} \quad f'(c) \text{ undefined}

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Critical points are the only candidates for local extrema.

THEOREMFermat's Theorem

If

f

has a local maximum or minimum at

c

, then

c

must be a critical point.

This is Fermat's theorem: a local extremum at an interior point requires the derivative to vanish or fail to exist.

Pitfall: Skipping points where the derivative is undefined

Solving only

f'(x) = 0

for

f(x) = x^{2/3}

finds nothing, since

f'(x) = \frac{2}{3}x^{-1/3}

is never

0

. But

f'(0)

does not exist, so

x = 0

is a critical point, and it is the minimum:

f(0) = 0

.

The converse is false. The function

f(x) = x^3

has

f'(0) = 0

, so

x = 0

is a critical point, but

f

has no extremum there—it increases through the origin. A critical point is a candidate that must be tested further.

Finding critical points is the first step in any extremum or optimization problem. The subsequent tests—first derivative test and second derivative test—classify what happens at each candidate.

x³ − 3x on [−3, 3]: critical points at x = ±1

The derivative 3x² − 3 vanishes at x = −1 and x = 1, and the tool marks both: a local maximum on the left, a local minimum on the right, each with a horizontal tangent. Nothing else in the interval is a candidate for an extreme except the two endpoints. Change the interval or the function and watch the candidates recomputed on the optimization and critical points visualizer.

Locating the critical points is the first step; classifying them is the work of the two tests that follow.

The First Derivative Test

The first derivative test classifies a critical point

c

by examining how

f'

changes sign around it.

If

f'

changes from positive to negative at

c

—the function rises then falls—then

f

has a local maximum at

c

. If

f'

changes from negative to positive—the function falls then rises—then

f

has a local minimum at

c

. If

f'

does not change sign—positive on both sides or negative on both sides—then

c

is neither a maximum nor a minimum.

First Derivative Test

f'(x) \text{ changes } + \to - \text{ at } c \implies c \text{ is a local max}

f'(x) \text{ changes } - \to + \text{ at } c \implies c \text{ is a local min}

Learn more about this formula: First Derivative Test →

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Fact: A minimum the sign test cannot see

f(x) = x^2\left(2 + \sin\frac{1}{x}\right)

, with

f(0) = 0

, has a strict minimum at

0

, since

f(x) \geq x^2

. Yet

f'(x) = 2x\left(2 + \sin\frac{1}{x}\right) - \cos\frac{1}{x}

takes both signs on every interval beside

0

, so the first derivative test has no sign to read.

The test works even when

f'(c)

does not exist, as long as

f

is continuous at

c

. It requires checking the sign of

f'

in an interval immediately to the left and immediately to the right of

c

. No information about

f''

is needed, which makes this test the most widely applicable of the derivative-based classification methods.

The test needs one sign of

f'

on a whole interval just left of

c

and one sign on a whole interval just right of it. For a non-constant polynomial such intervals always exist, because

f'

has only finitely many zeros and keeps one sign between neighboring ones. The test is a sufficient condition for an extremum: when it gives a verdict, the verdict is right.

A critical point that is neither max nor min

f(x) = x³ with f′(x) = 3x² and f″(x) = 6x. At x = 0 the derivative touches zero, so 0 is a critical point, but f′ is positive just to the left and just to the right: it does not change sign. By the first derivative test the point is neither a maximum nor a minimum, and the graph keeps rising through it. The second derivative is 0 there too, so only the sign of f′ settles the case. Try other families on the optimization and critical points visualizer.

Compare section 4’s x³ − 3x: there f′ crosses zero at ±1 and the test finds a maximum and a minimum.

The Second Derivative Test

When

f'(c) = 0

and

f''(c)

exists, the second derivative test provides a quicker classification.

If

f''(c) > 0

, the graph is concave up at

c

, and the horizontal tangent sits at the bottom of a cup:

f

has a local minimum at

c

. If

f''(c) < 0

, the graph is concave down, and the horizontal tangent sits at the top of a cap:

f

has a local maximum at

c

.

Second Derivative Test

f'(c) = 0, \; f''(c) > 0 \implies c \text{ is a local min}

f'(c) = 0, \; f''(c) < 0 \implies c \text{ is a local max}

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Pitfall: Swapping the two verdicts

Reading

f''(c) > 0

as a peak calls

x = 1

a maximum of

f(x) = x^3 - 3x

, where

f''(1) = 6

. A positive second derivative means a cup, so

x = 1

is the local minimum,

f(1) = -2

; the maximum is at

x = -1

, where

f''(-1) = -6

.

If

f''(c) = 0

, the test is inconclusive. The point might be an extremum or an inflection point—further analysis is required. In this case, fall back to the first derivative test or examine higher-order derivatives.

The second derivative test is faster than the first derivative test when

f''

is easy to compute, but it cannot handle critical points where

f'

does not exist. The table below sets the two tests side by side across their hypotheses, mechanics, and verdicts.

Test	When applicable	How it works	Verdict at c
First derivative test	f continuous at c (works even if f'(c) does not exist)	examine the sign of f' immediately to the left and right of c	+ → −: local maximum; − → +: local minimum; no sign change: neither
Second derivative test	f'(c) = 0 and f''(c) exists	evaluate the sign of f''(c)	f''(c) > 0: local minimum; f''(c) < 0: local maximum; f''(c) = 0: inconclusive

f″(c) = 2.60: concave up, a cup

At the marked point the curve bends upward and the second derivative is positive; a horizontal tangent here would sit at the bottom of the cup, which is the local-minimum case of the second derivative test. Turn the bend over and the sign flips with it. Move the marker to a downward bend on the concavity and inflection points tool.

When f″(c) = 0 the test is silent, and the first derivative test must decide.

Concavity

Concavity describes how the slope of

f

changes, not whether the function rises or falls but how it bends while doing so.

If

f''(x) > 0

on an interval, the derivative

f'

is increasing there: the slope gets steeper (or less negative). The graph bends upward—concave up. Visually, the curve lies above its tangent lines.

If

f''(x) < 0

on an interval, the derivative

f'

is decreasing: the slope gets less steep (or more negative). The graph bends downward—concave down. The curve lies below its tangent lines.

Concavity from Second Derivative

f''(x) > 0 \implies f \text{ concave up}, \quad f''(x) < 0 \implies f \text{ concave down}

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A function can be increasing and concave down simultaneously—rising but decelerating, like a ball thrown upward before reaching its peak. Concavity and direction are independent properties controlled by

f''

and

f'

respectively. The four sign combinations of

f'

and

f''

produce four qualitatively distinct shapes, collected in the table below alongside the visual picture each one produces.

Sign of f'	Sign of f''	What f does on the interval	Visual picture
+	+	increasing and concave up	rising with slope getting steeper
+	−	increasing and concave down	rising but flattening — like a thrown ball approaching its peak
−	+	decreasing and concave up	falling but flattening — approaching a local minimum from the left
−	−	decreasing and concave down	falling with slope getting more negative

f″(c) = −2.60: concave down

Here the slopes decrease as x increases: the tangent lines turn clockwise as they move right, and the curve stays below each of them. That is concavity read from the second derivative, negative throughout this stretch. Compare the two bends side by side on the concavity and inflection points tool.

Concavity is the shape of the shape: not where the graph goes, but how it curves on the way.

Inflection Points

An inflection point is a point where the concavity of

f

changes—from concave up to concave down, or the reverse.

Inflection Point Condition

f''(c) = 0 \text{ or undefined}, \text{ and } f''(x) \text{ changes sign at } c

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Fact: The bell curve bends one standard deviation out

For the standard bell curve

f(x) = e^{-x^2/2}

,

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

changes sign at

x = \pm 1

. So the curve turns from a cap to a cup exactly one standard deviation from its center, and

\sigma

can be read off a drawn bell curve as the distance to an inflection point.

At an inflection point,

f''

must either equal zero or fail to exist. However,

f''(c) = 0

alone does not guarantee an inflection point. The function

f(x) = x^4

has

f''(0) = 0

, but the concavity does not change at

x = 0

—the graph is concave up on both sides. The sign of

f''

must actually switch across the point.

To locate inflection points: find where

f''(x) = 0

or

f''

is undefined, then verify that

f''

changes sign. On the graph of

f'

, inflection points of

f

correspond to local extrema of

f'

—the slope of

f

reaches a peak or trough and reverses direction.

f″(c) = 0.00: the bending changes sides

To the left of the marker the curve cups down, to the right it cups up, and at the marker itself the second derivative is exactly zero while it changes sign: the definition of an inflection point in one picture. A zero of f″ without the sign change would not qualify. Slide the marker through the crossing on the concavity and inflection points tool.

Inflection points are to the second derivative what critical points are to the first.

Curve Sketching

Curve sketching assembles all derivative information into a complete picture of a function's graph.

The process begins with preliminary analysis: domain, intercepts, symmetry (even, odd, periodic), and asymptotes. Vertical asymptotes arise where the function is undefined; horizontal asymptotes come from limits at infinity.

First derivative analysis determines where

f

increases and decreases, and locates all local extrema. Second derivative analysis determines concavity intervals and inflection points. End behavior—the function's direction as

x \to \pm\infty

—frames the overall shape.

Combining these elements produces a qualitative sketch without plotting individual points. The derivative framework reveals structure that pointwise computation alone would miss: a function might appear flat in a table of values while the derivatives expose a subtle extremum nearby.

The signs of

f'

and

f''

, read interval by interval, fix the shape of a graph before a single point is plotted.

Optimization

Optimization finds the absolute maximum or minimum value of a function, often on a specified interval or subject to constraints.

On a closed interval

[a, b]

, the Extreme Value Theorem guarantees that a continuous function attains both an absolute maximum and an absolute minimum. The candidates are the critical points inside

(a, b)

and the endpoints

a

and

b

. Evaluate

f

at each candidate; the largest value is the absolute maximum, the smallest is the absolute minimum.

Extreme Value Theorem

f \text{ continuous on } [a, b] \implies f \text{ attains a max and min on } [a, b]

Learn more about this formula: Extreme Value Theorem →

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Pitfall: Forgetting the endpoints

Comparing only the critical values of

f(x) = x^3 - 3x

on

[-3, 3]

,

f(-1) = 2

and

f(1) = -2

, gives a maximum of

2

. The endpoints give

f(3) = 18

and

f(-3) = -18

, so the absolute maximum is

18

and the absolute minimum is

-18

.

Applied optimization problems translate a real scenario into a function of one variable. The steps are: identify the quantity to optimize, express it as a function of a single variable, determine the feasible domain, find critical points within that domain, and compare values at critical points and endpoints.

The derivative identifies candidates; the context determines which candidate solves the problem. Checking that a critical point is actually a maximum (or minimum) rather than merely a critical point is an essential final step. The table below collects this procedure as a five-step recipe.

Step	Action	Why it matters
1	identify the quantity to optimize	names the objective the derivative will be taken of
2	express it as a function of a single variable	constraints are used here to eliminate extra variables
3	determine the feasible domain	physical or geometric meaning restricts where the variable can live
4	find the critical points inside that domain	candidates for the extremum come from f'(x) = 0 or f'(x) undefined
5	compare values at critical points and at the endpoints	on a closed interval, the Extreme Value Theorem guarantees the maximum and minimum lie among these

x⁴ − 4x² on [−3, 3]: critical points and endpoints

The W-shaped quartic has three critical points, two minima at x = ±√2 and a local maximum at 0, and the tool lists them beside the two endpoints; the absolute extremes on the closed interval are found by comparing just these five values. That is the Extreme Value Theorem turned into a procedure. Narrow the interval and watch the absolute maximum move to an endpoint on the optimization and critical points visualizer.

Constraints change the function to be optimized, never the procedure.

Related Rates

Related rates problems involve two or more quantities that change simultaneously with respect to a common variable, usually time

t

. A known rate of change in one quantity determines an unknown rate of change in another, connected through an equation.

The method uses implicit differentiation with respect to

t

. Start with an equation relating the variables—geometric, physical, or algebraic. Differentiate both sides with respect to

t

, applying the chain rule to every variable that depends on

t

. Substitute known values and known rates, then solve for the unknown rate.

Pitfall: Dropping the chain-rule factor

Differentiating

A = \pi r^2

with respect to

t

as

\frac{dA}{dt} = 2\pi r

leaves out

\frac{dr}{dt}

. With

r = 5

and

\frac{dr}{dt} = 2

this gives

10\pi

; the chain rule gives

\frac{dA}{dt} = 2\pi r \frac{dr}{dt} = 20\pi

.

Setting up the relationship correctly is the critical step. The equation must hold at all times, not just at the particular instant in question. Constants can be substituted only after differentiation—replacing a variable with a fixed value before differentiating eliminates its rate of change from the equation.

Differentiate the relation that holds at every instant, and only then substitute the numbers of the moment.

The Full Derivative Toolkit at a Glance

Every section above introduced a separate way the first or second derivative answers a question about the graph of f — direction, steepness, extrema, concavity, inflection, sketching, optimization, related rates. The table below lays them all out side by side: each tool, the geometric or applied question it settles, and the specific derivative information it needs.

Tool	What it reveals	Derivative information needed
Value of f'(a)	slope and direction of f at the single point (a, f(a))	first derivative evaluated at a
Tangent line	best linear approximation to f near x = a: y − f(a) = f'(a)(x − a)	first derivative evaluated at a
Sign of f'	intervals where f is increasing or decreasing	first derivative across each interval
Critical points	candidates for local extrema (must be tested further)	solutions of f'(x) = 0 plus points where f' is undefined
First derivative test	classifies a critical point as max, min, or neither	sign of f' on each side of the critical point
Second derivative test	faster classification when f'' is available and nonzero	value of f''(c) at a critical point with f'(c) = 0
Sign of f''	intervals of concave up or concave down behavior	second derivative across each interval
Inflection points	points where concavity reverses	f''(x) = 0 or undefined, plus a sign change of f'' across the point
Curve sketching	a complete qualitative picture: monotonicity, extrema, concavity, asymptotes	f' and f'' combined with domain, intercepts, and end behavior
Optimization	absolute maximum or minimum on a domain or under a constraint	critical points of the objective function, plus endpoint values
Related rates	how an unknown rate of change depends on known rates of related quantities	implicit differentiation of a constraint equation with respect to time

Common Pitfalls

Six mistakes cause many wrong answers when derivatives are used to read a graph. Each is marked by a box where it happens in the sections above; here each is worked through in full, with the wrong and the right working side by side.

Leaving x in the slope

When the tangent line at

x = a

is written, the derivative goes into the point-slope formula as the function

f'(x)

instead of its value

f'(a)

. The slope then still contains

x

, and the equation is no longer a line.

Wrong working: for

f(x) = x^2

at

a = 3

,

f(3) = 9

and

f'(x) = 2x

, so

y - 9 = 2x(x - 3)

. This expands to

y = 2x^2 - 6x + 9

, a parabola.

Right working: evaluate the derivative first,

f'(3) = 6

. Then

y - 9 = 6(x - 3)

, so

y = 6x - 9

.

Why it happens: the formula

y - f(a) = f'(a)(x - a)

contains both

a

and

x

, and right after

f'(x)

has been computed the two letters are easy to merge.

How to check: a tangent line has the form

y = mx + b

with numbers

m

and

b

, and no power of

x

beyond the first. It must also pass through the point: at

x = 3

,

6 \cdot 3 - 9 = 9 = f(3)

.

Joining intervals across a break

f'

has the same sign on both sides of a point where

f

is undefined, and the two intervals are joined into one, as if

f

kept increasing across the gap.

Wrong working: for

f(x) = -\frac{1}{x}

,

f'(x) = \frac{1}{x^2} > 0

for every

x \neq 0

, so

f

is increasing on

(-\infty, 0) \cup (0, \infty)

.

Right working: the sign test speaks about one open interval at a time.

f

is increasing on

(-\infty, 0)

and increasing on

(0, \infty)

, but not on their union:

-1 < 1

, while

f(-1) = 1 > f(1) = -1

.

Why it happens: the hole at

x = 0

is easy to overlook when

f'

has the same sign on both sides of it, and a union of intervals looks like a tidy single answer.

How to check: take one input on each side of the break and compare the outputs. If the larger input gives the smaller output, the function is not increasing across the break.

Skipping points where the derivative is undefined

Critical points are found by solving

f'(x) = 0

only. Points of the domain where

f'

does not exist are left out, and an extremum at a corner or a cusp is lost.

Wrong working: for

f(x) = x^{2/3}

,

f'(x) = \frac{2}{3}x^{-1/3}

, which is never

0

. So

f

has no critical points and no extrema.

Right working:

f(0) = 0

is defined but

f'(0)

does not exist, so

x = 0

is a critical point. For

x < 0

,

f'(x) < 0

, and for

x > 0

,

f'(x) > 0

, so

f

has a local minimum at

0

. Since

x^{2/3} = (x^{1/3})^2 \geq 0

for every

x

, it is also the absolute minimum.

Why it happens: the condition

f'(x) = 0

turns into an equation to solve, while the condition "

f'(c)

does not exist" has nothing to solve and is easily forgotten.

How to check: after solving

f'(x) = 0

, look for values in the domain of

f

where the formula for

f'

breaks down: a zero denominator, a negative power of

x

, an absolute value.

Swapping the two verdicts

In the second derivative test, the sign of

f''(c)

is matched to the wrong verdict: a positive value is read as a peak and a negative value as a valley.

Wrong working: for

f(x) = x^3 - 3x

,

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

is

0

at

x = \pm 1

, and

f''(x) = 6x

. Since

f''(1) = 6 > 0

,

x = 1

is a local maximum.

Right working:

f''(1) > 0

means concave up, a cup, so

x = 1

is a local minimum,

f(1) = -2

. At

x = -1

,

f''(-1) = -6 < 0

, a cap, so

x = -1

is a local maximum,

f(-1) = 2

.

Why it happens: "positive" sounds like "high", and the maximum is the high point, so the positive sign gets attached to the maximum.

How to check: try the simplest case.

f(x) = x^2

has

f''(0) = 2 > 0

and an obvious minimum at

0

. In the example, the values agree:

f(1) = -2

is lower than

f(-1) = 2

.

Forgetting the endpoints

On a closed interval, the absolute extrema are chosen from the critical points alone, and the values at the endpoints are never computed.

Wrong working: for

f(x) = x^3 - 3x

on

[-3, 3]

, the critical points are

x = \pm 1

, with

f(-1) = 2

and

f(1) = -2

. So the absolute maximum is

2

and the absolute minimum is

-2

.

Right working: the candidates are the critical points and the endpoints.

f(-3) = -27 + 9 = -18

,

f(-1) = 2

,

f(1) = -2

,

f(3) = 27 - 9 = 18

. The absolute maximum is

18

at

x = 3

and the absolute minimum is

-18

at

x = -3

.

Why it happens: the derivative finds the points where the graph turns, and an endpoint is not a turning point, so it never appears when

f'(x) = 0

is solved.

How to check: a local maximum is only the highest point nearby. Evaluate

f

at both ends of the interval and compare before naming an absolute extremum.

Dropping the chain-rule factor

In a related rates problem, a quantity that depends on time is differentiated as if it were the variable of differentiation itself, and the factor from the chain rule is lost.

Wrong working: for a circle with area

A = \pi r^2

,

\frac{dA}{dt} = 2\pi r

. With

r = 5

cm and

\frac{dr}{dt} = 2

cm/s, this gives

10\pi

.

Right working:

r

depends on

t

, so

\frac{dA}{dt} = 2\pi r \frac{dr}{dt}

. At

r = 5

and

\frac{dr}{dt} = 2

:

\frac{dA}{dt} = 2\pi \cdot 5 \cdot 2 = 20\pi \approx 62.8

cm²/s.

Why it happens:

2\pi r

is the familiar derivative of

\pi r^2

with respect to

r

, and the switch from

r

to

t

as the variable is easy to miss.

How to check: look at the units.

2\pi r

is measured in cm, not in cm²/s. And if the radius stopped growing,

\frac{dr}{dt} = 0

, the area could not change, so the rate must contain

\frac{dr}{dt}

.

Graph Analysis FAQ

How do you find the equation of the normal line?

+

The normal line at x = a is perpendicular to the tangent there, so it passes through (a, f(a)) with slope −1/f′(a), giving y − f(a) = −(1/f′(a))(x − a). The negative reciprocal comes from perpendicularity. If f′(a) = 0 the tangent is horizontal, so the normal is the vertical line x = a and no slope formula applies.Read more →

Is every critical point a maximum or minimum?

+

No. Critical points are the only candidates for a local extremum, but many are not extrema. f(x) = x³ has f′(0) = 0, yet the function increases straight through the origin — no peak, no valley. That is why finding critical points is only the first step; the first or second derivative test has to classify each one.Read more →

Should you use the first or second derivative test?

+

Use the second derivative test when f′(c) = 0 and f″ is easy to compute — it is a single evaluation rather than a sign check on both sides. Use the first derivative test when f″(c) = 0, when f″ is awkward, or when f′(c) does not exist, since the second test cannot handle that case at all.Read more →

Can a function be increasing and concave down at the same time?

+

Yes — direction and bending are independent. The sign of f′ says whether the function rises or falls; the sign of f″ says how the slope is changing. A ball thrown upward is rising while decelerating: f′ > 0 and f″ < 0. The four sign combinations of f′ and f″ give four distinct graph shapes.Read more →

Does f″(x) = 0 always mean there is an inflection point?

+

No. It makes the point a candidate, not a verdict — the concavity has to actually reverse there. For f(x) = x⁴, f″(0) = 0, yet the graph is concave up on both sides of the origin, so there is no inflection. Always confirm that f″ changes sign across the point before calling it one.Read more →

Why can't you substitute values before differentiating in a related rates problem?

+

Because substituting a value first turns a variable into a constant, and constants have zero rate of change — the term you needed disappears from the differentiated equation. Set up a relation that holds at every instant, differentiate it with respect to t using implicit differentiation, and only then plug in the values true at that moment.Read more →

Graph Analysis with Derivatives

Reading Functions Through Their Derivatives

Key Terms

First Derivative

Second Derivative

The Derivative at a Point

Tangent Lines

Notation of Graph Analysis

Notation of Graph Analysis

Increasing and Decreasing Functions

Critical Points

The First Derivative Test

The Second Derivative Test

Concavity

Inflection Points

Curve Sketching

Optimization

Related Rates

The Full Derivative Toolkit at a Glance

Common Pitfalls

Graph Analysis FAQ

How do you find the equation of the normal line?

Is every critical point a maximum or minimum?

Should you use the first or second derivative test?

Can a function be increasing and concave down at the same time?

Does f″(x) = 0 always mean there is an inflection point?

Why can't you substitute values before differentiating in a related rates problem?