Second derivative test — concavity, inflection points: Exercises

4.1.29 · D4 · Maths › Calculus I — Limits & Derivatives › Second derivative test — concavity, inflection points

Definition Toolkit — is page par use hone wali har cheez, simple shabdon mein

$f (x)$ = position $x$ par road ki height.
$f^{'} (x)$ = slope = road kitni steep hai. $f^{'} > 0$ matlab upar chadh rahi hai, $f^{'} < 0$ matlab neeche ja rahi hai, $f^{'} = 0$ matlab momentarily flat hai.
$f^{''} (x)$ = slope ka derivative = steepness kaise badal rahi hai.
Ek critical point woh jagah hai jahan $f^{'} (x) = 0$ (wahan road flat hai). Peaks aur valleys ke candidates sirf yahin milte hain.
Concave up ( $\cup$ , ek "cup") ka matlab hai slope $f^{'}$ badh raha hai, jo exactly tab hota hai jab $f^{''} > 0$ — road upar ki taraf mudi hai aur paani rok sakti hai.
Concave down ( $\cap$ , ek "cap") ka matlab hai slope $f^{'}$ ghat raha hai, yani $f^{''} < 0$ — road neeche ki taraf mudi hai aur paani bahata hai.
Ek inflection point ek continuous jagah hai jahan concavity switch karti hai ( $\cup \leftrightarrow \cap$ ). Iske liye dono chahiye: $f^{''} = 0$ (ya $f^{''}$ undefined) AUR $f^{''}$ ka ek sachcha sign flip.

Neeche ki figure hamaari visual dictionary hai — baad ke har problem mein isi ki taraf refer kiya gaya hai. Iske parts ko seedha padho:

x-axis (horizontal) position $x$ hai; y-axis (vertical) height $f (x)$ hai.
Curve cubic $f (x) = x^{3} - 3 x$ hai.
Curve ka pink hissa (sab $x < 0$ ) concave down hai — ek cap $\cap$ — aur yeh local max (yellow dot) $x = - 1$ par carry karta hai, height $f (- 1) = 2$ .
Blue hissa (sab $x > 0$ ) concave up hai — ek cup $\cup$ — aur yeh local min (yellow dot) $x = 1$ par carry karta hai, height $f (1) = - 2$ .
$(0, 0)$ par yellow square inflection point hai: yeh woh jagah hai jahan $f^{''} = 6 x$ ka sign $-$ (cap) se $+$ (cup) mein badalta hai. Problem L3.1 is square ko poore sign check ke saath dobara derive karta hai.

Figure — Second derivative test — concavity, inflection points

Level 1 — Recognition

(Sirf $f^{''}$ ka sign padho aur concavity ka naam batao. Abhi koi calculus test nahi.)

L1.1 Ek point par jahan $f^{''} (x) = 7$ hai, graph concave up hai ya concave down?

Recall Solution L1.1

$f^{''} = 7 > 0$ . Sign kya kehta hai: slope $f^{'}$ badh raha hai. Yeh kaisa dikhta hai: road ek cup $\cup$ ki tarah bend karti hai. → Concave up.

L1.2 Ek point par jahan $f^{''} (x) = - 0.3$ hai, concavity batao aur yeh bhi batao ki agar wahan ek flat point ho ( $f^{'} = 0$ ) toh woh max hoga ya min.

Recall Solution L1.2

$f^{''} = - 0.3 < 0$ → slope ghat raha hai → cap $\cap$ → concave down. Ek flat point jo ek cap ke andar baitha ho, woh bump ke top par hota hai ⇒ local maximum.

L1.3 $f (x) = x^{2}$ ke liye, $f^{''}$ compute karo aur har jagah concavity batao.

Recall Solution L1.3

$f^{'} (x) = 2 x$ , isliye $f^{''} (x) = 2$ . Kyunki har $x$ ke liye $2 > 0$ hai, parabola har jagah concave up hai — ek akela tuta-nahi hua cup. Koi inflection point nahi hai (sirf $f^{''}$ ka sign kabhi nahi badlta).

Level 2 — Application

(Poora second-derivative test chalao.)

L2.1 $f (x) = x^{2} - 4 x + 1$ ke critical points classify karo.

Recall Solution L2.1

Step 1 (flat spots dhundho). $f^{'} (x) = 2 x - 4 = 0 \Rightarrow x = 2$ . Kyun: critical points woh hain jahan road flat ho. Step 2 (wahan concavity). $f^{''} (x) = 2$ . Kyun: test ko $x = 2$ par bend chahiye. Step 3 (verdict). $f^{''} (2) = 2 > 0 \Rightarrow$ $x = 2$ par local minimum. Value: $f (2) = 4 - 8 + 1 = - 3$ .

L2.2 $f (x) = - x^{2} + 6 x$ ke critical points classify karo.

Recall Solution L2.2

$f^{'} (x) = - 2 x + 6 = 0 \Rightarrow x = 3$ . $f^{''} (x) = - 2 < 0 \Rightarrow$ $x = 3$ par local maximum. Value $f (3) = - 9 + 18 = 9$ .

L2.3 $f (x) = x^{3} - 3 x^{2} + 4$ ke liye, saare critical points dhundho aur classify karo.

Recall Solution L2.3

Step 1. $f^{'} (x) = 3 x^{2} - 6 x = 3 x (x - 2) = 0 \Rightarrow x = 0$ ya $x = 2$ . Step 2. $f^{''} (x) = 6 x - 6$ . Step 3. $f^{''} (0) = - 6 < 0 \Rightarrow$ $x = 0$ par local max, value $f (0) = 4$ . $f^{''} (2) = 6 > 0 \Rightarrow$ $x = 2$ par local min, value $f (2) = 8 - 12 + 4 = 0$ .

Level 3 — Analysis

(Sign-changes chase karo, degenerate / inconclusive cases pakdo.)

L3.1 $f (x) = x^{3} - 3 x$ ke saare inflection points dhundho aur har ek ko sign check se confirm karo.

Recall Solution L3.1

$f^{''} (x) = 6 x$ . Candidate: $6 x = 0 \Rightarrow x = 0$ . Sign check (yahi toh poori baat hai): $x < 0$ ke liye, $f^{''} = 6 x < 0$ (cap $\cap$ ); $x > 0$ ke liye, $f^{''} > 0$ (cup $\cup$ ). Sign flip $- \to +$ ⇒ genuinely inflection at $(0, 0)$ , kyunki $f (0) = 0$ aur $f$ continuous hai. Yeh exactly page ke top par figure mein yellow square hai.

L3.2 Kya $f (x) = x^{4}$ ka $x = 0$ par inflection point hai? Apna reasoning dikhao.

Recall Solution L3.2

$f^{''} (x) = 12 x^{2}$ . Candidate: $12 x^{2} = 0 \Rightarrow x = 0$ . Lekin $x < 0$ ke liye, $12 x^{2} > 0$ ; $x > 0$ ke liye, $12 x^{2} > 0$ . Koi sign change nahi — dono sides concave up. Isliye koi inflection point nahi, chahe $f^{''} (0) = 0$ ho.

L3.3 $f (x) = x^{4} - 6 x^{2}$ ke liye, har inflection point dhundho.

Recall Solution L3.3

Step 1. $f^{'} (x) = 4 x^{3} - 12 x$ , $f^{''} (x) = 12 x^{2} - 12 = 12 (x^{2} - 1) = 12 (x - 1) (x + 1)$ . Step 2. Candidates: $f^{''} = 0 \Rightarrow x = \pm 1$ . Step 3 ( $12 (x - 1) (x + 1)$ ka sign map):

$x < - 1$ : dono factors negative → product $+$ → $\cup$ .
$- 1 < x < 1$ : $(x - 1) < 0, (x + 1) > 0$ → product $-$ → $\cap$ .
$x > 1$ : dono positive → $+$ → $\cup$ .

Sign flip dono $x = - 1$ aur $x = + 1$ par. Values: $f (\pm 1) = 1 - 6 = - 5$ . Inflections $(- 1, - 5)$ aur $(1, - 5)$ par.

L3.4 $f (x) = x^{1/3}$ kahan concave up / down hai, aur kya $x = 0$ par inflection hai?

Recall Solution L3.4

$f^{'} (x) = \frac{1}{3} x^{- 2/3}$ , $f^{''} (x) = - \frac{2}{9} x^{- 5/3}$ . $f^{''}$ $x = 0$ par exist nahi karta — lekin inflection ki definition mein " $f^{''} (c) = 0$ ya $f^{''} (c)$ undefined" dono allowed hain. Sign check: $x > 0$ ke liye, $x^{- 5/3} > 0$ toh $f^{''} < 0$ (cap $\cap$ ); $x < 0$ ke liye, $x^{- 5/3} < 0$ toh $f^{''} > 0$ (cup $\cup$ ). Sign flip $+ \to -$ across $0$ , aur $f$ continuous hai with $f (0) = 0$ . ⇒ $(0, 0)$ par inflection (vertical-tangent wala).

Level 4 — Synthesis

(Second-derivative test ko doosre tools ke saath combine karo: first-derivative test, curve sketching, Taylor.)

L4.1 $f (x) = x^{4} - 4 x^{3}$ ke liye, saare critical points dhundho, har ek classify karo (second-derivative test use karo, aur jahan woh inconclusive ho wahan first-derivative test), aur inflection points list karo.

Recall Solution L4.1

Critical points. $f^{'} (x) = 4 x^{3} - 12 x^{2} = 4 x^{2} (x - 3) = 0 \Rightarrow x = 0$ ya $x = 3$ . Second derivative. $f^{''} (x) = 12 x^{2} - 24 x = 12 x (x - 2)$ . $x = 3$ par: $f^{''} (3) = 12 \cdot 3 \cdot 1 = 36 > 0 \Rightarrow$ local min, $f (3) = 81 - 108 = - 27$ . $x = 0$ par: $f^{''} (0) = 0 \Rightarrow$ test inconclusive. First derivative test par fall back karo:

$0$ ke thoda left ( $x = - 1$ ): $f^{'} = 4 (1) (- 4) = - 16 < 0$ (downhill),
thoda right ( $x = 1$ ): $f^{'} = 4 (1) (- 2) = - 8 < 0$ (abhi bhi downhill).

Slope negative hi rehta hai $- \to -$ : $0$ par koi max ya min nahi — yeh ek saddle/shelf hai (ek flat spot jo abhi bhi neeche ja rahi road par hai). Inflection points. $f^{''} = 12 x (x - 2) = 0 \Rightarrow x = 0, 2$ . $12 x (x - 2)$ ka sign:

$x < 0$ : $(-) (-) = +$ → $\cup$ ; $0 < x < 2$ : $(+) (-) = -$ → $\cap$ ; $x > 2$ : $(+) (+) = +$ → $\cup$ .

Dono $x = 0$ aur $x = 2$ par flip. Toh $x = 0$ ek critical point hai jo inflection point bhi hai. Values $f (0) = 0$ , $f (2) = 16 - 32 = - 16$ . Inflections $(0, 0)$ aur $(2, - 16)$ par.

L4.2 Taylor series argument use karo yeh explain karne ke liye ki $f^{''} (c) = 0$ second-derivative test ko kyun fail karta hai, aur isse $x = 0$ par $f (x) = x^{4}$ ke critical point ko classify karo.

Recall Solution L4.2

Ek critical point $c$ (jahan $f^{'} (c) = 0$ ) ke paas, Taylor deta hai $f (x) = f (c) + \frac{1}{2} f^{''} (c) (x - c)^{2} + \frac{1}{6} f^{'''} (c) (x - c)^{3} + \frac{1}{24} f^{''''} (c) (x - c)^{4} + \dots$ Jab $f^{''} (c) \neq = 0$ hota hai, term $\frac{1}{2} f^{''} (c) (x - c)^{2}$ dominate karta hai aur $(x - c)^{2} \geq 0$ $f (x) - f (c)$ ke sign ko $f^{''} (c)$ se match karne pe majboor karta hai — clean verdict. Jab $f^{''} (c) = 0$ hota hai toh quadratic term vanish ho jaata hai, toh woh decide nahi kar sakta — exactly yahi wajah hai ki test inconclusive hai; agla non-zero term control leta hai. $f (x) = x^{4}$ ke liye $c = 0$ par: $f^{''} (0) = f^{'''} (0) = 0$ lekin $f^{''''} (0) = 24$ . Toh $f (x) - f (0) \approx \frac{1}{24} (24) x^{4} = x^{4} \geq 0$ . Kyunki $x^{4} \geq 0$ dono sides par hai, $f (x) \geq f (0)$ nearby ⇒ local minimum. (First-derivative test se match karta hai.)

L4.3 $f (x) = \frac{1}{1 + x ^{2}}$ ko sketch-classify karo: iska maximum aur uske inflection points dhundho. (Curve sketching ka prep.)

Recall Solution L4.3

First derivative. Quotient/chain rule use karte hue, $f^{'} (x) = \frac{- 2 x}{( 1 + x ^{2} ) ^{2}}$ . $= 0$ set karo $\Rightarrow x = 0$ . Second derivative. $f^{''} (x) = \frac{6 x ^{2} - 2}{( 1 + x ^{2} ) ^{3}}$ . $x = 0$ classify karo: $f^{''} (0) = \frac{- 2}{1} = - 2 < 0$ . Yeh peak kyun force karta hai: $f^{''} < 0$ ka matlab hai road flat spot $x = 0$ par hi downward bend kar rahi hai (ek cap $\cap$ ), toh $x = 0$ us downward bend ke top par baitha hai — har nearby point neeche hai. Isliye local (aur global) maximum, $f (0) = 1$ . Inflections. Numerator $6 x^{2} - 2 = 0 \Rightarrow x^{2} = \frac{1}{3} \Rightarrow x = \pm \frac{1}{3}$ . Denominator $(1 + x^{2})^{3} > 0$ hamesha, toh $f^{''}$ ka sign $6 x^{2} - 2$ ka sign hai:

$∣ x ∣ < \frac{1}{3}$ : numerator $< 0$ → $\cap$ ; $∣ x ∣ > \frac{1}{3}$ : numerator $> 0$ → $\cup$ .

Sign flip dono $\pm \frac{1}{3}$ par. Value $f (\pm \frac{1}{3}) = \frac{1}{1 + \frac{1}{3}} = \frac{3}{4}$ . Inflections $(\pm \frac{1}{3}, \frac{3}{4})$ par.

Level 5 — Mastery

(Poori generality mein reasoning: har case cover karo, ek claim prove karo, ek parameter handle karo.)

L5.1 Maano $f (x) = x^{3} + b x$ ek real parameter $b$ ke saath. $b$ ki har value ke liye determine karo ki $f$ ke kitne local extrema hain aur kya $x = 0$ ek inflection point hai.

Recall Solution L5.1

$f^{'} (x) = 3 x^{2} + b$ , $f^{''} (x) = 6 x$ . Critical points $3 x^{2} + b = 0 \Rightarrow x^{2} = - \frac{b}{3}$ solve karte hain.

Case $b > 0$ : $x^{2} = - b /3 < 0$ ka koi real solution nahi → koi critical points nahi → koi local extrema nahi. ( $f$ strictly increasing hai.)
Case $b = 0$ : $x^{2} = 0 \Rightarrow x = 0$ , lekin $f^{''} (0) = 0$ → test inconclusive. First-derivative test: $f^{'} = 3 x^{2} \geq 0$ har jagah, koi sign change nahi ⇒ shelf, koi extremum nahi.
Case $b < 0$ : $x = \pm - b /3$ , do critical points. $f^{''} (+ - b /3) > 0$ ⇒ min; $f^{''} (- - b /3) < 0$ ⇒ max. Toh exactly do extrema (ek min, ek max).

$x = 0$ par inflection: $f^{''} = 6 x$ har $b$ ke liye $0$ across $- \to +$ flip karta hai, aur $f$ continuous hai ⇒ $(0, 0)$ hamesha ek inflection point hai, $b$ chahe kuch bhi ho.

L5.2 Prove karo: agar $f^{''}$ continuous hai, $f^{''} (c) = 0$ hai, aur $f^{''}$ $c$ across sign change karta hai, toh $c$ ek inflection point hai; lekin ek concrete function deke dikhao ki " $f^{''} (c) = 0$ " akela sufficient nahi hai.

Recall Solution L5.2

Sufficiency ka proof. "Concave up" $⟺ f^{''} > 0$ ; "concave down" $⟺ f^{''} < 0$ . Agar $f^{''}$ $c$ par sign change karta hai (maano $c$ ke left par $-$ , right par $+$ ), toh $f$ $c$ ke immediately left par concave down hai aur immediately right par concave up hai. Inflection point ki definition se (ek continuous point jahan concavity change hoti hai), $c$ ek inflection point hai. $f$ ki continuity $c$ par hold karti hai kyunki $f^{''}$ (aur isliye $f^{'}$ aur $f$ ) wahan continuous hai. $■$ Necessity fail karti hai (" $f^{''} (c) = 0$ " akele): $f (x) = x^{4}$ lo. Toh $f^{''} (0) = 12 \cdot 0^{2} = 0$ phir bhi $f^{''} (x) = 12 x^{2} \geq 0$ kabhi sign change nahi karta → dono sides concave up → koi inflection nahi. Toh condition $f^{''} (c) = 0$ necessary-ish hai (ek candidate) lekin sufficient nahi.

L5.3 $f (x) = x^{5} - 5 x^{3}$ consider karo. Har critical point dhundho aur poori tarah classify karo (un sab ko bhi jo first-derivative test maangein), aur saare inflection points sign checks ke saath list karo.

Recall Solution L5.3

First derivative. $f^{'} (x) = 5 x^{4} - 15 x^{2} = 5 x^{2} (x^{2} - 3) = 0 \Rightarrow x = 0, x = \pm 3$ . Second derivative. $f^{''} (x) = 20 x^{3} - 30 x = 10 x (2 x^{2} - 3)$ . $x = 3$ par: $f^{''} (3) = 103 (2 \cdot 3 - 3) = 103 \cdot 3 = 303 > 0 \Rightarrow$ local min. $f (3) = (3)^{5} - 5 (3)^{3} = 93 - 153 = - 63$ . $x = - 3$ par: $f^{''} (- 3) = - 303 < 0 \Rightarrow$ local max, $f (- 3) = 63$ . $x = 0$ par: $f^{''} (0) = 0$ → inconclusive. $f^{'} = 5 x^{2} (x^{2} - 3)$ par first-derivative test: $0$ ke paas, $5 x^{2} > 0$ aur $(x^{2} - 3) < 0$ , toh $f^{'} < 0$ dono sides par ⇒ koi sign change nahi ⇒ shelf, extremum nahi. Inflection points. $f^{''} = 10 x (2 x^{2} - 3) = 0 \Rightarrow x = 0$ ya $x = \pm 3/2$ . Real line par $10 x (2 x^{2} - 3)$ ka sign (roots $- 1.5 \approx - 1.22, 0, + 1.5 \approx 1.22$ par):

$x < - 1.5$ : $(-) (+) = -$ → $\cap$
$- 1.5 < x < 0$ : $(-) (-) = +$ → $\cup$
$0 < x < 1.5$ : $(+) (-) = -$ → $\cap$
$x > 1.5$ : $(+) (+) = +$ → $\cup$

Teeno roots par sign flip ⇒ $x = 0, \pm 3/2$ par inflections. Values: $f (0) = 0$ ; $f (\pm 3/2) = \pm ((3/2)^{5/2} - 5 (3/2)^{3/2}) = \mp \frac{21}{4} \frac{3}{2}$ (numerically $\approx \mp 6.43$ ). Inflections $(0, 0), (3/2, - 6.43 \dots), (- 3/2, 6.43 \dots)$ par.

Recap ladder

Recall Har level ne kya train kiya

L1 $f^{''}$ ka sign padho → concavity. L2 Ek critical point par second-derivative test chalao. L3 $f^{''}$ sign-changes chase karo; $f^{''} = 0$ -no-flip aur $f^{''}$ -undefined inflections handle karo. L4 First-derivative test, Taylor, aur sketching ke saath combine karo. L5 Sign change ki sufficiency prove karo; saare parameter/boundary cases sweep karo.

Connections

Second Derivative Test — Concavity, Inflection Points
First derivative test
Critical points
Taylor series
Curve sketching
Maxima and minima — optimization
Concavity and convex functions