Second derivative test — concavity, inflection points

4.1.29 · Maths › Calculus I — Limits & Derivatives

1. Concavity kya hai? (slope-of-slope se build karo)

HOW detect karein "slope increasing hai"? Kisi bhi function $g$ ka "increasing" hona $g^{'} > 0$ se measure hota hai. Ise $g = f^{'}$ par apply karo:

$f^{'} increasing ⟺ (f^{'})^{'} > 0 ⟺ f^{''} (x) > 0.$

2. Inflection points

Figure — Second derivative test — concavity, inflection points

3. Second Derivative Test (max/min ke liye)

Taylor se derivation (first principles). $c$ ke paas, jahan $f^{'} (c) = 0$ : $f (x) = f (c) + f^{'} (c) (x - c) + \frac{1}{2} f^{''} (c) (x - c)^{2} + \dots = f (c) + \frac{1}{2} f^{''} (c) (x - c)^{2} + \dots$ Kyunki $(x - c)^{2} \geq 0$ hamesha hota hai, $f (x) - f (c)$ ka sign $c$ ke paas $f^{''} (c)$ ke sign se match karta hai:

$f^{''} (c) > 0 \Rightarrow f (x) > f (c)$ nearby ⇒ valley ⇒ min.
$f^{''} (c) < 0 \Rightarrow f (x) < f (c)$ nearby ⇒ peak ⇒ max.
$f^{''} (c) = 0 \Rightarrow$ quadratic term vanish ho jaata hai, higher terms decide karte hain ⇒ inconclusive. ∎

4. Worked examples

5. Active recall

Recall Khud test karo (attempt karne ke baad kholo)

Q1. $f^{''} > 0$ ka geometrically kya matlab hai? → Concave up ( $\cup$ ), slope increasing. Q2. $f^{'} (c) = 0$ aur $f^{''} (c) < 0$ : classify karo. → Local maximum. Q3. Kya $f^{''} (c) = 0$ inflection point ke liye kaafi hai? → Nahi; $f^{''}$ ko $c$ ke across sign change karna chahiye. Q4. Jab $f^{''} (c) = 0$ ho toh kya bachata hai? → First-derivative sign test. Q5. " $f^{''} = 0 \Rightarrow$ inflection" ka counterexample? → $x^{4}$ at $0$ .

Recall Feynman: ek 12-saal ke bacche ko samjhao

Socho tum ek pahadi road par drive kar rahe ho. Upar ya neeche jaana ek baat hai — lekin kya road smile ki tarah curve ho rahi hai ya frown ki tarah? Ek smile-curve ( $\cup$ ) matlab hai ki agar tum flat jagah ruko toh tum sabse neeche dip par ho (ek min). Ek frown-curve ( $\cap$ ) ki flat jagah ek bump ka top hai (ek max). Woh jagah jahan smile frown mein badlti hai woh inflection point hai — lekin tabhi jab woh sach mein switch kare. Kabhi kabhi road bas flatness ko kiss karti hai aur same curve rakhti hai (jaise $x^{4}$ ), isliye hamesha dono sides check karo.

6. 80/20 core

$f^{''}$ ka sign = concavity ( $+ \cup$ , $- \cap$ ).
$f^{'} = 0$ par: $f^{''} > 0$ min, $f^{''} < 0$ max, $f^{''} = 0$ inconclusive.
Inflection ⇔ $f^{''}$ sign change karta hai.

f^{''} (x) > 0

ka matlab graph... hai

Concave up (

\cup

); slope

f^{'}

increasing hai.

f^{''} (x) < 0

ka matlab graph... hai

Concave down (

\cap

); slope

f^{'}

decreasing hai.

Second derivative test jab

f^{'} (c) = 0, f^{''} (c) > 0

c

par local minimum.

Second derivative test jab

f^{'} (c) = 0, f^{''} (c) < 0

c

par local maximum.

Kya hota hai jab

f^{'} (c) = 0

aur

f^{''} (c) = 0

Test inconclusive hai; first-derivative sign test use karo.

Inflection point ki definition

Ek continuous point jahan concavity sign change karti hai (up↔down).

Kya

f^{''} (c) = 0

inflection point ke liye sufficient hai?

Nahi —

f^{''}

ko actually

c

ke across sign change karna chahiye (e.g.

x^{4}

fail karta hai).

f^{''} (c) > 0

min kyun deta hai (Taylor argument)?

f (x) - f (c) \approx \frac{1}{2} f^{''} (c) (x - c)^{2}

; kyunki

(x - c)^{2} \geq 0

, sign

f^{''} (c)

follow karta hai.

sin x

ka inflection point

(0, 2 π)

par

x = π

par, jahan

f^{''} = - sin x

sign change karta hai.

x^{3} - 3 x

ke critical points aur classification

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