Exercises — Applications — increasing - decreasing, local extrema (first derivative test)

Hum kya karte hain: test values ko $f^{\prime }$ mein plug karo aur sign dekho. Hume $f$ ki zarurat nahi — $f^{\prime }$ ka sign hi rising/falling ka poora map hai.

• $f^{\prime }(0)=(0-2)(0+3)=(-2)(3)=-6<0$ → $x=0$ par decreasing.
• $f^{\prime }(5)=(5-2)(5+3)=(3)(8)=24>0$ → $x=5$ par increasing.

Critical hone ke do tarike: $f^{\prime }=0$ ya $f^{\prime }$ undefined jabki $x$, $f$ ke domain mein ho.

• $f^{\prime }=0$: fraction zero tabhi hoga jab uska numerator zero ho → $x-1=0\Rightarrow x=1$. ✓
• $f^{\prime }$ undefined: $x=4$ par denominator zero hai. Lekin $x=4$ $f$ ke domain mein nahi hai, toh yeh $f$ ka critical point nahi hai — kisi point ko count hone ke liye domain mein hona zaruri hai.

Answer: sirf ek hi critical point hai: $x=1$.

Step 1 — differentiate. $f^{\prime }(x)=2x-6$ (power rule term by term; constant $+5$ ki slope $0$ hai). Step 2 — critical points. $2x-6=0\Rightarrow x=3$. $f^{\prime }$ ek polynomial hai, har jagah defined, toh koi "undefined" points nahi. Step 3–4 — test points $x=3$ ke dono taraf:

• $x=0$: $f^{\prime }(0)=-6<0$ → $(-\infty ,3)$ par decreasing.
• $x=4$: $f^{\prime }(4)=2>0$ → $(3,\infty )$ par increasing. Step 5 — classify. Sign $-\to +$ jaata hai → $x=3$ par local minimum. Uski value hai $f(3)=9-18+5=-4$. Answer: $(-\infty ,3)$ par decreasing, $(3,\infty )$ par increasing; local min at $(3,-4)$.

Step 1. $f^{\prime }(x)=6x^2-6x-12$. Step 2. Zero set karo: $6x^2-6x-12=0\Rightarrow x^2-x-2=0\Rightarrow (x-2)(x+1)=0\Rightarrow x=-1,2$. Step 3–4. Teen intervals; har ek mein ek point test karo (yahi exactly top figure ka $+,-,+$ chart hai):

• $x=-2$: $f^{\prime }=6(4)-6(-2)-12=24+12-12=24>0$ → $(-\infty ,-1)$ par increasing.
• $x=0$: $f^{\prime }=-12<0$ → $(-1,2)$ par decreasing.
• $x=3$: $f^{\prime }=6(9)-18-12=54-30=24>0$ → $(2,\infty )$ par increasing. Step 5.
• At $x=-1$: $+\to -$ → local max, value $f(-1)=2(-1)-3(1)-12(-1)+1=-2-3+12+1=8$.
• At $x=2$: $-\to +$ → local min, value $f(2)=2(8)-3(4)-24+1=16-12-24+1=-19$. Answer: local max $(-1,8)$, local min $(2,-19)$.

Step 1. $f^{\prime }(x)=\frac{2}{3}{x}^{-1/3}=\frac{2}{3\sqrt[3]{x}}$. Step 2 — critical points. $f^{\prime }=0$ ka koi solution nahi (numerator $2/3$ kabhi zero nahi hota). Lekin $f^{\prime }$ at $x=0$ undefined hai ($\sqrt[3]{0}=0$ se division), aur $x=0$ $f$ ke domain mein hai (kyunki $0^{2/3}=0$). Toh $x=0$ ek critical point hai — "derivative-does-not-exist" wale kind ka. Step 3–4. Dono taraf test karo. Key hai $\sqrt[3]{x}$ ka sign: yeh $x<0$ ke liye negative aur $x>0$ ke liye positive hota hai.

• $x=-1$: $\sqrt[3]{-1}=-1$, toh $f^{\prime }=\frac{2}{3(-1)}=-\frac{2}{3}<0$ → $(-\infty ,0)$ par decreasing.
• $x=1$: $\sqrt[3]{1}=1$, toh $f^{\prime }=\frac{2}{3}>0$ → $(0,\infty )$ par increasing. Step 5. Sign $-\to +$ → $x=0$ par local minimum, value $f(0)=0$. Geometrically yeh ek sharp cusp hai — graph neeche aata hai, point karta hai, aur wapas upar jaata hai. Answer: $(0,0)$ par local min; koi aur critical point nahi.

Part A — $f(x)=x^4$ (honest minimum). $f^{\prime }(x)=4x^3$. $4x^3=0$ set karne par ek hi critical point milta hai $x=0$. Dono taraf test karo:

• $x=-1$: $f^{\prime }=4(-1{)}^3=-4<0$ → decreasing.
• $x=1$: $f^{\prime }=4(1{)}^3=4>0$ → increasing. Sign $-\to +$ → $(0,0)$ par local minimum. Toh $x^4$ genuinely ek extremum rakhta hai — yeh "zero slope ⇒ extremum" ka counterexample nahi hai, bas yeh ek aisa case hai jahan implication hold karta hai.

Part B — $g(x)=(x-1{)}^5$ (real trap). $g^{\prime }(x)=5(x-1{)}^4$ (chain rule: outer power $5$, inner derivative $1$). Zero karne par $x=1$ milta hai, ek critical point. Kyunki $(x-1{)}^4\ge 0$ hamesha (even power), $g^{\prime }\ge 0$ har jagah:

• $x=0$: $g^{\prime }=5(-1{)}^4=5>0$ → increasing.
• $x=2$: $g^{\prime }=5(1{)}^4=5>0$ → increasing. Sign $+\to +$ — dono taraf same sign → koi extremum nahi. Point $(1,0)$ horizontal tangent wala ek inflection point hai. Answer: $x^4$ par $0$ par local min hai; $g=(x-1{)}^5$ par $x=1$ par critical point hai lekin koi extremum nahi. Fark: $f^{\prime }=4x^3$ at $0$ sign change karta hai, jabki $g^{\prime }=5(x-1{)}^4$ (even power) nahi karta.

Pehle domain. Hume $4-x^2\ge 0\Rightarrow -2\le x\le 2$ chahiye. Toh domain hai closed interval $[-2,2]$. Step 1 — differentiate (product + chain rule). $f=x(4-x^2{)}^{1/2}$ likho. $f^{\prime }(x)=1\cdot (4-x^2{)}^{1/2}+x\cdot \frac{1}{2}(4-x^2{)}^{-1/2}(-2x)=\sqrt{4-x^2}-\frac{x^2}{\sqrt{4-x^2}}.$ Ek denominator par combine karo: $f^{\prime }(x)=\frac{(4-x^2)-x^2}{\sqrt{4-x^2}}=\frac{4-2x^2}{\sqrt{4-x^2}}.$ Step 2 — critical points.

• $f^{\prime }=0$: numerator zero → $4-2x^2=0\Rightarrow x^2=2\Rightarrow x=\pm \sqrt{2}$. Dono $(-2,2)$ ke andar hain. ✓
• $f^{\prime }$ undefined: denominator $x=\pm 2$ par zero — lekin yeh domain ke endpoints hain, toh inhe alag neeche one-sided reasoning se classify karenge. Step 3–4 — $f^{\prime }$ ka sign (denominator $\sqrt{4-x^2}>0$ interior par, toh sign numerator $4-2x^2$ follow karta hai):
• $x=-1.5$: $4-2(2.25)=-0.5<0$ → $(-2,-\sqrt{2})$ par decreasing.
• $x=0$: $4-0=4>0$ → $(-\sqrt{2},\sqrt{2})$ par increasing.
• $x=1.5$: $-0.5<0$ → $(\sqrt{2},2)$ par decreasing. Step 5 — interior critical points classify karo.
• At $x=-\sqrt{2}$: $-\to +$ → local min. Value $f(-\sqrt{2})=-\sqrt{2}\cdot \sqrt{4-2}=-\sqrt{2}\cdot \sqrt{2}=-2$.
• At $x=\sqrt{2}$: $+\to -$ → local max. Value $f(\sqrt{2})=\sqrt{2}\cdot \sqrt{2}=2$. Step 6 — endpoints classify karo (one-sided test). Closed domain par ek endpoint local extremum hota hai agar $f$ "downhill" taraf move kare jab tum usse andar ki taraf step karo.
• Left endpoint $x=-2$ par: thoda right mein, $(-2,-\sqrt{2})$ par, $f^{\prime }<0$ (decreasing) hai. Toh $x=-2$ se andar step karne par function neeche jaata hai, matlab $x=-2$ apne immediate neighbours se upar hai → local maximum. Value $f(-2)=-2\cdot \sqrt{4-4}=-2\cdot 0=0$.
• Right endpoint $x=2$ par: thoda left mein, $(\sqrt{2},2)$ par, $f^{\prime }<0$, toh $x=2$ ke paas function abhi bhi gir raha hai; is se $x=2$ apne immediate neighbours se neeche hai → local minimum. Value $f(2)=2\cdot 0=0$. Answer: interior local min $(-\sqrt{2},-2)$ aur local max $(\sqrt{2},2)$; endpoint local max $(-2,0)$ aur endpoint local min $(2,0)$.

Domain. $x^2+1$ kabhi zero nahi → domain saare real $x$ hain. Step 1 — quotient rule. $f^{\prime }(x)=\frac{(1)(x^2+1)-x(2x)}{(x^2+1{)}^2}=\frac{x^2+1-2x^2}{(x^2+1{)}^2}=\frac{1-x^2}{(x^2+1{)}^2}.$ Step 2. Denominator $(x^2+1{)}^2>0$ hamesha, toh $f^{\prime }=0\iff 1-x^2=0\Rightarrow x=\pm 1$. Koi undefined points nahi. Step 3–4 — sign numerator $1-x^2$ follow karta hai:

• $x=-2$: $1-4=-3<0$ → $(-\infty ,-1)$ par decreasing.
• $x=0$: $1>0$ → $(-1,1)$ par increasing.
• $x=2$: $-3<0$ → $(1,\infty )$ par decreasing. Step 5.
• At $x=-1$: $-\to +$ → local min, value $f(-1)=\frac{-1}{2}=-\frac{1}{2}$.
• At $x=1$: $+\to -$ → local max, value $f(1)=\frac{1}{2}=\frac{1}{2}$. Answer: local min $(-1,-\frac{1}{2})$, local max $(1,\frac{1}{2})$.

Step 1. $f^{\prime }(x)=3x^2+a$. Reasoning. Local extremum ke liye ek critical point chahiye jahan $f^{\prime }$ ka sign change ho. Kyunki $f^{\prime }=3x^2+a$ $x$ mein ek upward parabola hai:

• Agar $a>0$: $3x^2+a>0$ sabhi $x$ ke liye (kabhi zero nahi) → $f^{\prime }$ kabhi zero nahi, $f$ strictly increasing → koi extrema nahi. ✓
• Agar $a=0$: $f^{\prime }=3x^2\ge 0$, sirf $x=0$ par zero lekin $+\to +$ (sign change nahi) → koi extrema nahi (inflection). ✓
• Agar $a<0$: $3x^2+a=0$ ke do roots hain $x=\pm \sqrt{-a/3}$, aur parabola unke beech zero se neeche jaata hai → $f^{\prime }$ $+,-,+$ jaata hai → genuine local max phir local min → extrema hain. Answer: $f$ ka koi local extrema nahi exactly tab jab $a\ge 0$.

Goal. Dikhao: $I$ mein kisi bhi $a<b$ ke liye, $f(a)<f(b)$ (increasing ki definition). Setup. Kyunki $f$, $I$ par differentiable hai (iska matlab hai "$f^{\prime }(x)$ exist karta hai"), yeh $[a,b]$ par continuous aur $(a,b)$ par differentiable bhi hai — exactly Mean Value Theorem ke hypotheses. MVT apply karo. Koi $c\in (a,b)$ exist karta hai jiske saath $f^{\prime }(c)=\frac{f(b)-f(a)}{b-a}.$ MVT kyun aur kuch nahi? Hume slope $f^{\prime }$ ke baare mein ek statement ko values $f(b)$ vs $f(a)$ ke baare mein statement mein convert karna hai. MVT exactly woh bridge hai: yeh $[a,b]$ par average slope ko andar kisi jagah actual instantaneous slope ke barabar karta hai. Finish. Rearrange karo: $f(b)-f(a)=f^{\prime }(c)(b-a)$. Hypothesis se $f^{\prime }(c)>0$, aur $b-a>0$ kyunki $a<b$. Positive times positive positive hota hai, toh $f(b)-f(a)>0$, yaani $f(b)>f(a)$. Kyunki $a<b$ arbitrary the, $f$, $I$ par increasing hai. $■$

Step 1. $f^{\prime }(x)=\cos x-\sin x$. Step 2 — interior critical points. $\cos x=\sin x\Rightarrow \tan x=1$. $\cos x$ se divide karne se pehle, check karo ki woh points jahan $\cos x=0$ hai (yaani $x=\frac{\pi }{2},\frac{3\pi }{2}$) miss na ho jayein:

• $x=\frac{\pi }{2}$ par: $f^{\prime }=\cos \frac{\pi }{2}-\sin \frac{\pi }{2}=0-1=-1\ne 0$.
• $x=\frac{3\pi }{2}$ par: $f^{\prime }=\cos \frac{3\pi }{2}-\sin \frac{3\pi }{2}=0-(-1)=1\ne 0$. Dono critical nahi hain, toh divide karna safe tha. $[0,2\pi ]$ par, $\tan x=1$ deta hai $x=\frac{\pi }{4}$ aur $x=\frac{5\pi }{4}$. Step 3–4 — $f^{\prime }=\cos x-\sin x$ ka sign:
• $x=0$: $1-0=1>0$ → $(0,\frac{\pi }{4})$ par increasing.
• $x=\pi$: $-1-0=-1<0$ → $(\frac{\pi }{4},\frac{5\pi }{4})$ par decreasing.
• $x=2\pi$: $1-0=1>0$ → $(\frac{5\pi }{4},2\pi )$ par increasing. Step 5 — interior critical points.
• At $x=\frac{\pi }{4}$: $+\to -$ → local max, value $\sin \frac{\pi }{4}+\cos \frac{\pi }{4}=\frac{\sqrt{2}}{2}+\frac{\sqrt{2}}{2}=\sqrt{2}$.
• At $x=\frac{5\pi }{4}$: $-\to +$ → local min, value $-\frac{\sqrt{2}}{2}-\frac{\sqrt{2}}{2}=-\sqrt{2}$. Step 6 — endpoints (one-sided test).
• Left endpoint $x=0$: thoda right mein, $(0,\frac{\pi }{4})$ par, $f^{\prime }>0$ (increasing) hai. Andar step karne par function upar jaata hai, toh $x=0$ apne neighbours se neeche hai → local minimum, value $f(0)=\sin 0+\cos 0=0+1=1$.
• Right endpoint $x=2\pi$: thoda left mein, $(\frac{5\pi }{4},2\pi )$ par, $f^{\prime }>0$ (increasing) hai, toh $2\pi$ ke paas pahunchte waqt function abhi bhi chadh raha hai; is se $x=2\pi$ apne neighbours se upar hai → local maximum, value $f(2\pi )=\sin 2\pi +\cos 2\pi =0+1=1$. Answer: interior local max $\left(\frac{\pi }{4},\sqrt{2}\right)$, interior local min $\left(\frac{5\pi }{4},-\sqrt{2}\right)$; endpoint local min $(0,1)$ aur endpoint local max $(2\pi ,1)$.