Connecting a Function, Its First Derivative, and Its Second Derivative

AP Calculus · Analytical Applications of Differentiation

Connecting $f$, $f'$ and $f''$

Move fluently between the graph, the table, the formula and the sentence — they all describe the same behaviour.

key idea$f'\ \text{gives direction};\quad f''\ \text{gives concavity}$

Four views of one idea

The CED asks you to move between the graphical, numerical, analytical and verbal descriptions of the same function — not just to compute. Here is one function, $f(x)=x^{3}-3x^{2}+2$, shown four ways. Every fact in one panel is visible in the others.

Graphical

[Figure: graph of f with max, inflection and min marked]

Numerical

$x$	$f$	$f'$	$f''$
$-1$	$-2$	$9$	$-12$
$0$	$2$	$0$	$-6$
$1$	$0$	$-3$	$0$
$2$	$-2$	$0$	$6$
$3$	$2$	$9$	$12$

Analytical

$f(x)=x^{3}-3x^{2}+2$
$f'(x)=3x^{2}-6x=3x(x-2)$
$f''(x)=6x-6=6(x-1)$

So $f'=0$ at $x=0,2$ and $f''=0$ at $x=1$.

Verbal

$f$ rises to a peak at $x=0$, falls to a valley at $x=2$, then rises again. All the way down it is bending — first curving downward, then, from $x=1$ onward, curving upward. The steepest descent happens exactly at $x=1$.

Translating between views

Stack the three graphs and the correspondence becomes mechanical. A zero of $f'$ sits under an extremum of $f$; a zero of $f''$ sits under an inflection point of $f$ and under the extremum of $f'$.

[Figure: stacked graphs of f, f prime and f double prime]

One function, three graphs. Dashed guides at $x=0,1,2$ line up the features: $f'=0$ under each extremum of $f$, $f''=0$ under the inflection point.

sign chart — $f'(x)=3x(x-2)$ and $f''(x)=6(x-1)$

	$x\lt 0$	$0\lt x\lt 1$	$1\lt x\lt 2$	$x\gt 2$
$f'$	+	−	−	+
$f''$	−	−	+	+
verdict	rising, concave down	falling, concave down	falling, concave up	rising, concave up

If you see this in $f'$ / $f''$	…it means this about $f$	…and looks like this
$f'\gt 0$	$f$ is increasing	graph rises left→right
$f'$ changes $+\to-$	local maximum of $f$	a peak
$f'$ changes $-\to+$	local minimum of $f$	a valley
$f'=0$ with no sign change	neither — a flat spot	a shelf, e.g. $x^{3}$ at $0$
$f''\gt 0$	$f$ is concave up	holds water
$f''\lt 0$	$f$ is concave down	sheds water
$f''$ changes sign	inflection point of $f$	bending switches direction
$f'$ has a local min	$f$ is steepest downward there	$f''=0$ at that $x$

The right-hand column is the verbal translation the CED asks you to produce.

Worked examples

EX 1

Full analysis in one pass

$f(x)=x^{3}-3x^{2}+2$

Describe every interval of increase/decrease and concavity, and classify all extrema and inflection points.

Differentiate twice

$f'(x)=3x^{2}-6x=3x(x-2)$ and $f''(x)=6x-6=6(x-1)$.

Find the candidate $x$-values

$f'=0$ at $x=0$ and $x=2$. $f''=0$ at $x=1$. These three values cut the line into the four intervals in the sign chart above.

Read $f'$ for direction

$f'\gt 0$ on $(-\infty,0)$ and $(2,\infty)$; $f'\lt 0$ on $(0,2)$. So $f$ increases, decreases, then increases.

Classify the extrema

$f'$ goes $+\to-$ at $x=0$ ⇒ local max at $(0,2)$. $f'$ goes $-\to+$ at $x=2$ ⇒ local min at $(2,-2)$.

Read $f''$ for concavity

$f''\lt 0$ for $x\lt 1$ (concave down), $f''\gt 0$ for $x\gt 1$ (concave up). The sign genuinely changes, so $(1,0)$ is an inflection point.

State it verbally

$f$ rises and bends downward until it peaks at $x=0$; keeps falling, still bending downward, until $x=1$; then keeps falling but now bending upward; bottoms out at $x=2$ and rises thereafter.

EX 2

Given only the graph of $f'$

graph → conclusion

The middle panel of the stacked figure shows $f'$. Without ever seeing $f$, say where $f$ has its extrema and where it is concave up.

Extrema come from zeros of $f'$

$f'$ crosses zero at $x=0$ and $x=2$. Those are the only candidates.

Use the sign change, not the zero

$f'$ is above the axis before $0$ and below after ⇒ max at $x=0$. Below before $2$ and above after ⇒ min at $x=2$.

Concavity is the SLOPE of $f'$

$f$ is concave up exactly where $f'$ is increasing. From the panel, $f'$ decreases until $x=1$ and increases after ⇒ $f$ is concave up on $(1,\infty)$.

Sanity check against $f''$

$f''=6(x-1)$ is positive for $x\gt 1$. Agrees.

EX 3

Given only a table

numerical → conclusion

A differentiable $g$ has $g'(1)=-4$, $g'(3)=0$, $g'(5)=6$, and $g''\gt 0$ on $[1,5]$. What happens at $x=3$?

Read the direction change

$g'$ moves from $-4$ to $0$ to $6$: it passes from negative to positive at $x=3$.

Apply the First Derivative Test

A $-\to+$ change in $g'$ at a critical point gives a local minimum at $x=3$.

Cross-check with the Second Derivative Test

$g''(3)\gt 0$ is given, which independently confirms a local minimum.

Note what the table cannot tell you

Nothing here locates an inflection point: $g''\gt 0$ throughout, so concavity never switches on $[1,5]$.

EX 4

Spot the contradiction

verbal → check

A student writes: “$h$ is decreasing and concave up on $(0,4)$, and $h$ has a local maximum at $x=2$.” What is wrong?

Translate each clause

Decreasing on $(0,4)$ means $h'\lt 0$ on all of $(0,4)$. Concave up means $h''\gt 0$ there.

Test the third clause

A local maximum at $x=2$ needs $h'$ to change from $+$ to $-$ at $x=2$. But $h'\lt 0$ on the whole interval — it never turns positive.

Name the contradiction

A function that is strictly decreasing on an interval has no interior local extremum on that interval. The first and third clauses cannot both hold.

Repair it

Either drop the extremum, or weaken “decreasing” to “decreasing then increasing.” Concave up is fine either way — concavity says nothing about direction.

Watch out

“Increasing at a decreasing rate” is two statements

Parse it as direction and bending: increasing means $f'\gt 0$; at a decreasing rate means $f'$ is falling, i.e. $f''\lt 0$. Concave down while still going up.

$f''=0$ is not enough for an inflection point

The sign of $f''$ must actually change. For $f(x)=x^{4}$, $f''(0)=0$ but $f''\ge 0$ everywhere — no inflection at the origin.

Don't apply one derivative's test to the other

Extrema come from $f'$; concavity and inflection come from $f''$. A zero of $f''$ says nothing about maxima or minima.

Reading the wrong graph

If the picture you are handed is $f'$, then its $y$-values are slopes of $f$, not heights of $f$. A peak on the graph of $f'$ is where $f$ is steepest, not where $f$ is largest.

Course map. Topic 5.9 · LO FUN-4.A — justify conclusions about the behaviour of $f$ from the behaviour of $f'$ and $f''$ (EK FUN-4.A.11: key features of the graphs of $f$, $f'$ and $f''$ are related to one another).