Second derivative test — concavity, inflection points

1. What is concavity? (build it from slope-of-slope)

HOW do we detect "slope is increasing"? "Increasing" of any function $g$ is measured by $g'>0$ . Apply this to $g=f'$ :

$f' \text{ increasing} \iff (f')' > 0 \iff f''(x) > 0.$

2. Inflection points

Figure — Second derivative test — concavity, inflection points

3. The Second Derivative Test (for max/min)

Derivation from Taylor (first principles). Near $c$ , with $f'(c)=0$ : $f(x) = f(c) + f'(c)(x-c) + \tfrac{1}{2}f''(c)(x-c)^2 + \cdots = f(c) + \tfrac12 f''(c)(x-c)^2 + \cdots$ Since $(x-c)^2 \ge 0$ always, the sign of $f(x)-f(c)$ near $c$ matches the sign of $f''(c)$ :

$f''(c)>0 \Rightarrow f(x) > f(c)$ nearby ⇒ valley ⇒ min.
$f''(c)<0 \Rightarrow f(x) < f(c)$ nearby ⇒ peak ⇒ max.
$f''(c)=0 \Rightarrow$ quadratic term vanishes, higher terms decide ⇒ inconclusive. ∎

4. Worked examples

5. Active recall

Recall Test yourself (open after attempting)

Q1. What does $f''>0$ mean geometrically? → Concave up ( $\cup$ ), slope increasing. Q2. $f'(c)=0$ and $f''(c)<0$ : classify. → Local maximum. Q3. Is $f''(c)=0$ enough for an inflection point? → No; $f''$ must change sign across $c$ . Q4. What rescues you when $f''(c)=0$ ? → First-derivative sign test. Q5. Counterexample to " $f''=0 \Rightarrow$ inflection"? → $x^4$ at $0$ .

Recall Feynman: explain to a 12-year-old

Imagine driving on a hilly road. Going up or down is one thing — but is the road curving like a smile or a frown? A smile-curve ( $\cup$ ) means if you stop on a flat bit you'll be at the lowest dip (a min). A frown-curve ( $\cap$ ) flat bit is the top of a bump (a max). The spot where the smile turns into a frown is the inflection point — but only if it truly switches. Sometimes the road just kisses flatness and keeps the same curve (like $x^4$ ), so always check both sides.

6. The 80/20 core

Sign of $f''$ = concavity ( $+\,\cup$ , $-\,\cap$ ).
At $f'=0$ : $f''>0$ min, $f''<0$ max, $f''=0$ inconclusive.
Inflection ⇔ $f''$ changes sign.

f''(x)>0

means the graph is...

Concave up (

\cup

); slope

f'

is increasing.

f''(x)<0

means the graph is...

Concave down (

\cap

); slope

f'

is decreasing.

Second derivative test when

f'(c)=0,\ f''(c)>0

Local minimum at

c

Second derivative test when

f'(c)=0,\ f''(c)<0

Local maximum at

c

What happens when

f'(c)=0

and

f''(c)=0

Test is inconclusive; use first-derivative sign test.

Definition of an inflection point

A continuous point where concavity changes sign (up↔down).

f''(c)=0

sufficient for an inflection point?

No —

f''

must actually change sign across

c

(e.g.

x^4

fails).

Why does

f''(c)>0

give a min (Taylor argument)?

f(x)-f(c)\approx\tfrac12 f''(c)(x-c)^2

; since

(x-c)^2\ge0

, sign follows

f''(c)

Inflection point of

\sin x

(0,2\pi)

x=\pi

, where

f''=-\sin x

changes sign.

Critical points & classification of

x^3-3x

Max at

x=-1

(

f=2

), min at

x=1

(

f=-2

), inflection at

(0,0)

Connections

First derivative test
Critical points
Taylor series
Curve sketching
Maxima and minima — optimization
Concavity and convex functions

Concept Map

Hinglish (regional understanding)

Intuition Hinglish mein samjho

Dekho, derivative ka funda simple hai. Pehla derivative $f'(x)$ batata hai road upar ja raha hai ya neeche (slope). Lekin doosra derivative $f''(x)$ batata hai ki road kaise mud raha hai — smile ki tarah (cup, $\cup$ , concave up) ya frown ki tarah (cap, $\cap$ , concave down). Yeh isliye kaam karta hai kyunki $f''$ asal me "slope ka slope" hai — agar slope badh raha hai to curve upar mudega, aur slope badhne ka matlab $f''>0$ .

Ab second derivative test ka magic: jahan $f'(c)=0$ (tangent flat hai), wahan agar $f''(c)>0$ hai matlab cup ke bottom me baithe ho → minimum. Agar $f''(c)<0$ hai matlab cap ke top pe ho → maximum. Yaad rakhne ka trick: "Up is a cuP (plus → min), down is a caP (minus → max)". Taylor expansion se proof aata hai: $f(x)-f(c)\approx \tfrac12 f''(c)(x-c)^2$ , aur $(x-c)^2$ hamesha positive, to sign sirf $f''(c)$ pe depend karta hai.

Inflection point wahan hota hai jahan concavity badalti hai — cup se cap ya cap se cup. Yahan ek bada trap hai: sirf $f''(c)=0$ hone se inflection nahi banta! $x^4$ dekho — $f''=12x^2$ zero hota hai $0$ pe, par sign nahi badalta, dono taraf concave up. To rule yeh hai: $f''=0$ sirf candidate deta hai, aapko check karna padega ki sign actually flip ho raha hai ya nahi.

Aur jab $f''(c)=0$ aa jaaye critical point pe, ghabrao mat — test fail ho gaya, to purana first derivative sign test use karo. Exam me yahi 80/20 hai: $f''$ ka sign = concavity, flat point pe sign = max/min, aur inflection = sign flip.

Second derivative test — concavity, inflection points

1. What is concavity? (build it from slope-of-slope)

2. Inflection points

3. The Second Derivative Test (for max/min)

4. Worked examples

5. Active recall

6. The 80/20 core

Connections

Concept Map

Hinglish (regional understanding)

Go deeper — visual, from zero

Test yourself — Calculus I — Limits & Derivatives

Connections