Using the First Derivative Test to Determine Relative (Local) Extrema

AP Calculus · Analytical Applications of Differentiation

Using the First Derivative Test

Not where $f'$ is zero — where $f'$ changes sign.

key idea$+\!\rightarrow\!-\ \Rightarrow\ \text{max};\qquad -\!\rightarrow\!+\ \Rightarrow\ \text{min}$

Reading the graph

The First Derivative Test does not ask where $f'$ is zero. It asks where $f'$ changes sign. Those are different questions, and the gap between them is where most marks are lost. Work from a sign chart, not from a list of critical points.

[Figure: graph of f prime crossing the axis at x=-1 and touching it at x=2]

Here $f'(x)=(x+1)(x-2)^{2}$. There are two critical points, but only one extremum: at $x=-1$ the graph of $f'$ crosses the axis; at $x=2$ it merely touches it and stays positive.

sign chart — $f'(x)=(x+1)(x-2)^{2}$

	$x\lt -1$	$-1\lt x\lt 2$	$x\gt 2$
$(x+1)$	−	+	+
$(x-2)^{2}$	+	+	+
$f'$	−	+	+
verdict	$f$ decreasing	$f$ increasing	$f$ increasing

At a critical point, $f'$ goes…	Conclusion about $f$	Why
$+ \rightarrow -$	local maximum	$f$ climbs, then falls — the top of a hill
$- \rightarrow +$	local minimum	$f$ falls, then climbs — the bottom of a valley
$+ \rightarrow +$ or $- \rightarrow -$	no extremum	direction never reverses — a flat shelf
undefined, with a sign change	extremum can still occur	e.g. a cusp; $f'$ need not exist at the extremum
undefined, no sign change	no extremum	e.g. a vertical tangent on a rising curve

An even-power factor such as $(x-2)^{2}$ never changes sign — that is exactly why $x=2$ above fails to produce an extremum.

Critical pointA value $c$ in the domain where $f'(c)=0$ or $f'(c)$ is undefined. A candidate for an extremum, nothing more.

Stationary pointA critical point where $f'(c)=0$ specifically. Every stationary point is critical; a cusp is critical but not stationary.

Local extremumA point where $f$ is largest or smallest compared with nearby points. Requires $f'$ to change sign.

Sign change$f'$ passes from positive to negative or negative to positive. This — not the zero itself — is what creates an extremum.

Worked examples

EX 1

Apply the test end to end

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

Find and classify all local extrema of a function whose derivative is $f'(x)=(x+1)(x-2)^{2}$.

Find the critical points

$f'=0$ when $x=-1$ or $x=2$. Both are critical points; $f'$ is defined everywhere, so there are no others.

Build the sign chart

Test one point in each interval. $f'(-2)=(-1)(16)=-16\lt 0$; $f'(0)=(1)(4)=4\gt 0$; $f'(3)=(4)(1)=4\gt 0$.

Classify $x=-1$

$f'$ goes $-\to+$, so $f$ has a local minimum at $x=-1$.

Classify $x=2$

$f'$ goes $+\to+$ — no sign change. So there is no extremum at $x=2$, despite $f'(2)=0$.

Answer

One local minimum, at $x=-1$. The point $x=2$ is a stationary point of inflection, not an extremum.

EX 2

No sign change, no extremum

$f(x)=x^{3}$

Explain why $f(x)=x^{3}$ has no local extremum at $x=0$ even though $f'(0)=0$.

Differentiate

$f'(x)=3x^{2}$, which is zero at $x=0$, so $x=0$ is a critical point.

Test the sign on both sides

$f'(-1)=3\gt 0$ and $f'(1)=3\gt 0$.

Apply the test

$f'$ is positive on both sides — $+\to+$. The function is increasing through $x=0$; it merely flattens momentarily.

Conclude

No extremum. “$f'(c)=0$” makes $c$ a candidate, nothing more.

EX 3

From the graph of $f'$ alone

graph → conclusion

The figure above shows $f'$. Without any formula for $f$, state the intervals where $f$ increases and classify every extremum.

Read where $f'$ is above/below the axis

$f'\lt 0$ for $x\lt -1$; $f'\gt 0$ for $x\gt -1$ (touching zero at $x=2$ without crossing).

Translate to $f$

$f$ decreases on $(-\infty,-1)$ and increases on $(-1,\infty)$.

Classify

The only axis crossing is at $x=-1$, going $-\to+$: a local minimum.

Watch the touch point

At $x=2$ the curve returns to the axis and turns back up. Touching is not crossing, so no extremum there.

EX 4

When $f'$ does not exist

cusp

Let $f(x)=x^{2/3}$. Classify the behaviour at $x=0$.

Differentiate

$f'(x)=\tfrac{2}{3}x^{-1/3}=\dfrac{2}{3\sqrt[3]{x}}$, which is undefined at $x=0$.

Still a critical point

A critical point is where $f'$ is zero or undefined, provided $f$ itself is defined there. $f(0)=0$, so $x=0$ qualifies.

Test the signs

For $x\lt 0$ the cube root is negative, so $f'\lt 0$. For $x\gt 0$, $f'\gt 0$.

Classify

$f'$ goes $-\to+$, so $f$ has a local minimum at $x=0$ — a cusp. The derivative never exists there, and the extremum is real all the same.

Watch out

A zero derivative is not an extremum

$f'(c)=0$ only nominates $c$. Without a sign change there is no extremum — Examples 1 and 2 both turn on this.

Even-power factors never change sign

Factors like $(x-2)^{2}$ or $(x+3)^{4}$ contribute a zero but no sign flip. Spot them in the factored derivative and you can predict the non-extrema before testing.

Do not forget where $f'$ is undefined

Cusps and corners are critical points too. Restricting attention to $f'=0$ silently drops them, as in Example 4.

Local, not global

The First Derivative Test classifies local behaviour. For absolute extrema on a closed interval you must also test the endpoints.

Course map. Topic 5.4 · LO FUN-4.A — justify conclusions about the behaviour of a function based on the behaviour of its derivatives (EK FUN-4.A.3).