Using the First Derivative Test to Determine Relative (Local) Extrema
Introduced| English |
|---|
| First Derivative Test/fɜːst dɪˈrɪvətɪv test/ |
| relative (local) maximum/ˈrelətɪv ˈmæksɪməm/ |
| relative (local) minimum/ˈrelətɪv ˈmɪnɪməm/ |
Classify a critical point by how $f'$ turns
- A critical point is a candidate extremum — now we decide which kind it is.
- The First Derivative Test 一阶导数判别法 reads the sign change of $f'$ across the point.
- Uphill-then-downhill makes a peak; downhill-then-uphill makes a valley.
- It reuses your sign chart from the last lesson — no new computation.
Positive → negative means a max
- If $f'$ changes from positive to negative at a critical point $c$, then $f$ has a relative (local) maximum 局部最大值 there.
- The function was climbing, levels off, then falls — the top of a hill.
- Read it straight off the sign chart: a $+$ to the left of $c$, a $-$ to the right.
- The maximum value is $f(c)$; the location is $x=c$.
A max then a min
y = ax³ + cx
The left hump is a local max ($f'$ goes $+$ to $-$) and the right dip a local min ($-$ to $+$).
If $f'$ changes from positive to negative at $c$, then $f$ has a...
Up then down = the top of a hill = local max.
Negative → positive means a min
- If $f'$ changes from negative to positive at $c$, then $f$ has a relative (local) minimum 局部最小值 there.
- Falling, levels off, then climbing — the bottom of a valley.
- Sign chart: a $-$ left of $c$, a $+$ right of $c$.
- If $f'$ does not change sign at $c$, there is no extremum there (a flat spot).
If $f'$ changes from negative to positive at $c$, then $f$ has a local ____ there.
Down then up = bottom of a valley = local min.
The routine
- Find critical points, build the sign chart of $f'$, and read each crossing.
- $+\to-$: local max. $-\to+$: local min. No change: neither.
- Works even where $f'$ is undefined (a corner), as long as the sign flips.
- This is the AP's most-used tool for classifying extrema.
If $f'$ has the same sign on both sides of a critical point, there is no extremum there.
No sign change (like $y=x^3$ at $0$) means no max or min.
For $f(x)=x^3-3x$, the critical point $x=1$ is a...
$f'$ goes $-\to+$ at $x=1$ → local min.
For $f(x)=x^3-3x$, the local max is at $x=-1$. Find the max value $f(-1)$.
$f(-1)=(-1)^3-3(-1)=-1+3=2$.
Select all true statements about the First Derivative Test.
It is the change of sign that matters, not the value at $c$.
It's the change in sign that matters, not the value of $f'$ at $c$ (which is $0$ or undefined there). If $f'$ keeps the same sign on both sides — like $y=x^3$ at $0$ (positive on both sides) — there is no extremum, even though it's a critical point.
Classify the critical points of $f(x)=x^3-3x$ ($f'=3(x-1)(x+1)$).
- At $x=-1$: $f'$ goes $+\to-$ (test $-2$ gives $+$, test $0$ gives $-$) → local max.
- At $x=1$: $f'$ goes $-\to+$ (test $0$ gives $-$, test $2$ gives $+$) → local min.
- Values: local max $f(-1)=2$, local min $f(1)=-2$.
The First Derivative Test classifies a critical point by the sign change of $f'$: $+\to-$ is a local maximum, $-\to+$ is a local minimum, and no sign change means no extremum. Just read the crossings off your $f'$ sign chart.