[Figure: FiveStudy logo]
FiveStudy
Calculus AB · Unit 5 · 5.7
AP Calculus · Analytical Applications of Differentiation
Using the Second Derivative Test
At a horizontal tangent, the bending decides. One substitution replaces a sign chart — when it works.
key idea$f'(c)=0\text{ and }f''(c)\gt 0\Rightarrow\text{min};\quad f''(c)\lt 0\Rightarrow\text{max}$
Reading the graph
At a point where the tangent is horizontal, the bending decides which kind of extremum you have. Concave up at a flat spot means the bottom of a bowl; concave down means the top of a hill. One substitution replaces an entire sign chart — when it works.
$f''\gt 0$ at the critical point → minimum
$f''\lt 0$ at the critical point → maximum
the test at a critical point $c$
| $f''(c)\gt 0$ | $f''(c)\lt 0$ | $f''(c)=0$ |
|---|
| conclusion | local min | local max | no conclusion |
| verdict | concave up at $c$ | concave down at $c$ | fall back to the first derivative |
scope and limits
- The test needs $f'(c)=0$ — it says nothing at a cusp or corner.
- $f''(c)=0$ is inconclusive, not evidence of anything.
- It classifies local extrema only; endpoints are outside its scope.
- When it fails, the First Derivative Test always works.
Worked examples
EX 1
One substitution, done
$f(x)=x^{3}-3x$Classify the critical points of $f(x)=x^{3}-3x$ using the Second Derivative Test.
1
Find the critical points
$f'(x)=3x^{2}-3=0$ gives $x=\pm1$.
2
Compute the second derivative
$f''(x)=6x$.
3
Test $x=-1$
$f''(-1)=-6\lt 0$, so $f$ is concave down there ⇒ local maximum.
4
Test $x=1$
$f''(1)=6\gt 0$, so concave up ⇒ local minimum.
5
Compare with the alternative
The First Derivative Test would need three sign tests. Here two substitutions settled it.
EX 2
When the test is silent
$f(x)=x^{4}$Apply the Second Derivative Test to $f(x)=x^{4}$ at $x=0$.
1
Confirm it is critical
$f'(x)=4x^{3}$, so $f'(0)=0$. The test may be attempted.
2
Compute $f''$
$f''(x)=12x^{2}$, so $f''(0)=0$.
3
Read the verdict
$f''(c)=0$ means the test is inconclusive — it gives no information at all.
4
Fall back
First Derivative Test: $f'(-1)=-4\lt 0$ and $f'(1)=4\gt 0$, a $-\to+$ change.
5
Conclude
Local minimum at $x=0$. The second-derivative test failing does not mean there is no extremum — only that this tool cannot see it.
EX 3
Choosing between the tests
$f'$ hard to signFor $f(x)=xe^{-x}$, classify the critical point.
1
Find the critical point
$f'(x)=e^{-x}-xe^{-x}=e^{-x}(1-x)$, zero at $x=1$.
2
Consider the first-derivative route
$e^{-x}\gt 0$ always, so the sign of $f'$ follows $(1-x)$: positive then negative. That works.
3
Try the second-derivative route
$f''(x)=e^{-x}(x-2)$, so $f''(1)=e^{-1}(-1)\lt 0$.
4
Conclude
Local maximum at $x=1$, by either route.
5
Which to prefer
Use the second-derivative test when $f''$ is easy to compute at one point. Use the first when $f''$ is messy or vanishes.
EX 4
The test does not apply
cuspFor $f(x)=x^{2/3}$, can the Second Derivative Test classify the behaviour at $x=0$?
1
Check the precondition
$f'(x)=\dfrac{2}{3x^{1/3}}$ is undefined at $x=0$ — it is not zero there.
2
State the consequence
The test requires $f'(c)=0$. That hypothesis fails, so the test cannot be used at all.
3
Note $f''$ is no help either
$f''(x)=-\tfrac29 x^{-5/3}$ is also undefined at $0$.
4
Use the first derivative instead
$f'\lt 0$ for $x\lt 0$ and $f'\gt 0$ for $x\gt 0$: a $-\to+$ change.
5
Conclude
Local minimum at $x=0$, a cusp. The First Derivative Test handles critical points the second cannot reach.
Watch out
$f''(c)=0$ is not evidence
It is a failure of the test, not a finding. $x^{4}$ has a genuine minimum where $f''=0$; $x^{3}$ has none. Both give $f''=0$.
The test needs a horizontal tangent
If $f'(c)$ is undefined rather than zero, the test does not apply — Example 4.
Local only
It classifies local behaviour. Absolute extrema on a closed interval still require checking the endpoints.
The first-derivative test never fails
It works at every critical point, including cusps and the $f''=0$ cases. The second is a shortcut, not a replacement.