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

Learning objectives

—FUN-4.A—Justify conclusions about the behavior of a function based on the behavior of its derivatives—Topic 5.4• FUN-4.A Justify conclusions about the behavior of a function based on the behavior of its derivatives.• FUN-4.A.2 The first derivative of a function can determine the location of relative (local) extrema of the function.• Enduring understanding FUN-4: A function’s derivative can be used to understand some behaviors of the function.

The First Derivative Test classifies a critical point by tracking how $f$ moves on either side. Because $f'(x)>0$ means $f$ is increasing and $f'(x)<0$ means $f$ is decreasing, a local extremum occurs only when the sign of $f'$ changes.

Sign of $f'$ through $x=c$	Behavior of $f$	Conclusion at $c$
$+\to-$	increasing, then decreasing	local maximum
$-\to+$	decreasing, then increasing	local minimum
$+\to+$ or $-\to-$	same direction on both sides	neither

Find the critical points in the domain, use them to split the domain into intervals, and determine the sign of $f'$ on each interval. Then state the sign change and the corresponding conclusion; the sign chart is the justification.

For $f(x)=x^3-3x$ , $f'(x)=3(x-1)(x+1),$ so the critical points are $x=-1$ and $x=1$ . The derivative signs are positive on $(-\infty,-1)$ , negative on $(-1,1)$ , and positive on $(1,\infty)$ . Therefore $f'$ changes $+\to-$ at $x=-1$ , giving a local maximum, and $-\to+$ at $x=1$ , giving a local minimum.

A critical point is only a candidate. The equation $f'(c)=0$ alone does not prove a local extremum; the First Derivative Test requires signs on both sides of $c$ .

Next Topic

—5.5 Using the Candidates Test to Determine Absolute (Global) ExtremaRead