Exercises — Curve sketching — systematic approach

Domain. Ek fraction mein sirf yahi cheez problem kar sakti hai — zero se divide karna. Denominator ko zero set karo: $x-3=0\Rightarrow x=3$. Toh function har real number ke liye defined hai siwaaye $3$ ke: $\text{Domain}=\mathbb{R}\setminus \{3\}.$ Vertical asymptote. $x=3$ par denominator $\to 0$ ho jaata hai jabki numerator $\to 3+2=5\ne 0$ rehta hai. Ek nonzero number ko kisi bahut chhoti cheez se divide karo toh woh explode karta hai, isliye ${\lim }_{x\to 3^{\pm }}f(x)=\pm \infty \Rightarrow \text{vertical asymptote at }x=3.$ Numerator $\ne 0$ kyun matter karta hai: agar dono top aur bottom vanish ho jaate, toh hamare paas ek wall ki jagah removable hole ho sakti thi — numerator hamesha check karo. Dekho Rational functions and asymptotes.

Har $x$ ki jagah $-x$ daalo aur simplify karo: $g(-x)=(-x{)}^4-3(-x{)}^2+1=x^4-3x^2+1=g(x).$ Kyunki $(-x{)}^{\text{even power}}=x^{\text{even power}}$, kuch bhi nahi badla. Kyunki $g(-x)=g(x)$ hai, function even hai — iska graph $y$-axis ke across ek mirror image hai. (Sirf $x$ ke even powers appear hote hain, jo ek even polynomial ki quick fingerprint hai.)

Horizontal asymptote yeh poochta hai: jab $x$ infinity ki taraf jaata hai toh curve kahan settle hota hai? Highest power $x^2$ se top aur bottom dono divide karo: $h(x)=\frac{3+\frac{1}{x^2}}{1+\frac{5}{x^2}}.$ Jab $x\to \pm \infty$, toh terms $\frac{1}{x^2}$ aur $\frac{5}{x^2}$ shrink hokar $0$ ho jaate hain (ek fixed number ek huge number ke upar). Toh ${\lim }_{x\to \pm \infty }h(x)=\frac{3+0}{1+0}=3\Rightarrow y=3.$ Yeh Limits and continuity ke limit idea ko use karta hai.

$y$-intercept: $x=0$ plug karo: $f(0)=0-0=0$. Point $(0,0)$. $x$-intercepts: $f(x)=0$ solve karo. Pehle factor karo — common $x$ bahar nikalo: $x^3-4x=x(x^2-4)=x(x-2)(x+2)=0.$ Product tab zero hota hai jab ek factor zero ho, toh $x=0,2,-2$. Intercepts $(-2,0),(0,0),(2,0)$ par hain.

Differentiate karo. $f^{\prime }(x)=3x^2-6x=3x(x-2).$ Critical points wahan hain jahan $f^{\prime }=0$ (yahan $f^{\prime }$ kabhi undefined nahi hai): $3x(x-2)=0\Rightarrow x=0,2$. $f^{\prime }$ ka sign chart (har region mein ek number test karo — dekho First derivative and monotonicity):

• $x<0$, maano $x=-1$: $3(-1)(-3)=+9>0$ → rising.
• $0<x<2$, maano $x=1$: $3(1)(-1)=-3<0$ → falling.
• $x>2$, maano $x=3$: $3(3)(1)=+9>0$ → rising.

$x=0$ par: $+\to -$ ⇒ local maximum, value $f(0)=0$ → $(0,0)$. $x=2$ par: $-\to +$ ⇒ local minimum, value $f(2)=8-12=-4$ → $(2,-4)$.

Upar se, $f^{\prime }(x)=3x^2-6x$. Dobara differentiate karo: $f^{\prime \prime }(x)=6x-6=6(x-1).$ $f^{\prime \prime }$ yeh measure karta hai ki slope khud rise kar raha hai ya fall (Second derivative test).

• $x<1$: $f^{\prime \prime }<0$ → concave down ($⌢$).
• $x>1$: $f^{\prime \prime }>0$ → concave up ($⌣$).

Concavity $x=1$ par flip hoti hai, aur wahan $f^{\prime \prime }$ sign change karta hai, toh $(1,f(1))=(1,1-3)=(1,-2)$ ek inflection point hai.

$f^{\prime }(x)=3x^2$, jo $x=0$ par $0$ hai — ek genuine critical point. Ab dono sides par $f^{\prime }$ ka sign test karo:

• $x<0$: $3x^2>0$ (ek square positive hota hai).
• $x>0$: $3x^2>0$.

Slope dono sides par positive hai — woh kabhi sign nahi badlti — toh First Derivative Test ke hisaab se koi local extremum nahi hai. Baaki check karo $f^{\prime \prime }(x)=6x$: woh $x<0$ ke liye $-$ hai, $x>0$ ke liye $+$ hai, toh concavity $0$ par flip hoti hai. Isse $(0,0)$ ek flat (horizontal) inflection point banta hai: tangent level hai, phir bhi curve $⌢$ se $⌣$ mein switch karte hue chadhta rehta hai.

Figure kya dikhata hai: blue curve $x^3$ origin se guzarti hui draw ki gayi hai; yellow dashed segment woh horizontal tangent hai (slope $0$) jo $(0,0)$ par touch kar raha hai. Origin ke left mein curve ko concave down ($⌢$, red) label kiya gaya hai, right mein concave up ($⌣$, green). Notice karo ki flat tangent ke bawajood, curve left-to-right rise karta rehta hai — yahi visual proof hai ki zero slope peak ya valley nahi hoti.

Numerator ki degree ($2$) denominator ki degree ($1$) se exactly ek zyada hai — yahi slant asymptote ka signal hai. Division karo: $f(x)=\frac{x^2+1}{x}=x+\frac{1}{x}.$ Bacha hua $\frac{1}{x}\to 0$ jab $x\to \pm \infty$, toh curve is line ko hug karta hai: $y=x.$ Kaise approach karta hai: bade positive $x$ ke liye, $\frac{1}{x}>0$, toh curve $y=x$ ke thoda upar hota hai. Bade negative $x$ ke liye, $\frac{1}{x}<0$, toh woh thoda neeche hota hai. Woh opposite sides se line ki taraf squeeze karta hai — kabhi touch nahi karta, kyunki $\frac{1}{x}\ne 0$.

Domain. Denominator $x^2+1$ hamesha $\ge 1>0$ hota hai (ek square plus ek kabhi zero nahi ho sakta). Toh $f$ saare real $x$ ke liye defined hai: domain $=\mathbb{R}$. Koi vertical asymptote nahi: vertical asymptotes ke liye denominator ka $0$ hona zaroori hai; woh kabhi nahi hota, toh koi nahi hai. Pehle domain check karne ka yahi faayda hai. Symmetry. $f(-x)=\frac{(-x{)}^2}{(-x{)}^2+1}=\frac{x^2}{x^2+1}=f(x)$ → even. Horizontal asymptote. $x^2$ se divide karo: $f(x)=\frac{1}{1+1/x^2}\to \frac{1}{1+0}=1$ jab $x\to \pm \infty$, toh $y=1$. Kyunki $x^2<x^2+1$ hamesha, $f(x)<1$ har jagah — curve $y=1$ ki taraf neeche se approach karta hai aur kabhi nahi pahunchta.

1. Domain. $x^2-4=(x-2)(x+2)=0$ at $x=\pm 2$: domain $\mathbb{R}\setminus \{-2,2\}$. 2. Intercepts. $f(0)=\frac{1}{-4}=-\frac{1}{4}$ → $(0,-\frac{1}{4})$. $f=0$ set karo: ek fraction tab zero hota hai jab uska numerator zero ho, lekin numerator constant $1\ne 0$ hai, toh koi $x$-intercept nahi. 3. Symmetry. $f(-x)=\frac{1}{x^2-4}=f(x)$ → even, $y$-axis ke upar mirror. 4. Asymptotes. Vertical $x=\pm 2$ par (denominator $\to 0$, numerator $=1\ne 0$). Horizontal: $\frac{1}{x^2-4}\to 0$ jab $x\to \pm \infty$ (constant over huge number), toh $y=0$. 5. First derivative. $f=(x^2-4{)}^{-1}$ likho aur chain rule use karo: $f^{\prime }(x)=-1\cdot (x^2-4{)}^{-2}\cdot 2x=\frac{-2x}{(x^2-4{)}^2}.$ Denominator $>0$ har jagah jahan defined hai. Numerator $-2x=0$ at $x=0$. Sign: $f^{\prime }>0$ for $x<0$, $f^{\prime }<0$ for $x>0$ ⇒ local maximum at $(0,-\frac{1}{4})$. 6. Second derivative aur concavity. $f^{\prime }=-2x(x^2-4{)}^{-2}$ ko product + chain rules se differentiate karo: $f^{\prime \prime }(x)=-2(x^2-4{)}^{-2}+(-2x)\cdot (-2)(x^2-4{)}^{-3}\cdot 2x=\frac{-2(x^2-4)+8x^2}{(x^2-4{)}^3}=\frac{6x^2+8}{(x^2-4{)}^3}.$ Numerator $6x^2+8$ hamesha positive hai (ek square-term aur $8$ ka sum). Toh $f^{\prime \prime }$ ka sign denominator $(x^2-4{)}^3$ ka sign hai:

• $|x|>2$: $x^2-4>0\Rightarrow (x^2-4{)}^3>0\Rightarrow f^{\prime \prime }>0$ → concave up ($⌣$).
• $|x|<2$: $x^2-4<0\Rightarrow (x^2-4{)}^3<0\Rightarrow f^{\prime \prime }<0$ → concave down ($⌢$).

Sign kyun flip hota hai: numerator kabhi sign nahi badlta, toh sirf denominator hi $f^{\prime \prime }$ flip kar sakta hai, aur woh exactly wahan flip karta hai jahan $x^2-4$ zero cross karta hai — $x=\pm 2$ par. Lekin woh points domain mein nahi hain, toh koi inflection point nahi hai: concavity ek wall ke across change hoti hai, curve par kisi point par nahi. 7. Assemble. Teen pieces: ek left branch ($x<-2$ ke liye) jo $+\infty$ se $0$ ki taraf dive karta hai aur concave up hai; ek middle hump (concave down) jo $(0,-\frac{1}{4})$ par peak karta hai aur dono walls par $-\infty$ ki taraf fall karta hai; ek right branch ($x>2$ ke liye) jo left ko mirror karta hai, concave up.

Figure kya dikhata hai: $1/(x^2-4)$ ki teen blue branches; do red dashed vertical lines walls $x=\pm 2$ hain; yellow dashed horizontal line floor $y=0$ hai. Green dot local maximum $(0,-\frac{1}{4})$ ko middle hump par mark karta hai — note karo ki woh $x$-axis ke neeche hai phir bhi apni branch ka highest point hai. Dekho kaise har outer branch $y=0$ ki taraf flatten hoti hai (concave up) jabki middle branch downward cap karti hai.

Domain: saara $\mathbb{R}$ (exponential har jagah defined hai). Intercepts: $f(0)=0\cdot e^0=0$. $f(x)=0$ ke liye $x=0$ chahiye (kyunki $e^{-x}$ kabhi $0$ nahi hota). Sirf $(0,0)$. First derivative (product rule): $f^{\prime }(x)=1\cdot e^{-x}+x\cdot (-e^{-x})=e^{-x}(1-x).$ $e^{-x}>0$ hamesha, toh sign $(1-x)$ ka hai: $f^{\prime }>0$ for $x<1$, $f^{\prime }<0$ for $x>1$ ⇒ $+\to -$ ⇒ local maximum at $x=1$, height $f(1)=1\cdot e^{-1}=\frac{1}{e}\approx 0.368$. Second derivative: $f^{\prime \prime }(x)=\frac{d}{dx}[e^{-x}(1-x)]=-e^{-x}(1-x)+e^{-x}(-1)=e^{-x}(x-2).$ Sign $(x-2)$ ka hai: $x<2$ ke liye concave down, $x>2$ ke liye concave up ⇒ inflection at $x=2$, height $f(2)=2e^{-2}\approx 0.271$. End behaviour. Jab $x\to +\infty$, $e^{-x}$ iss speed se decay karta hai jo $x$ ke growth se kahin zyada fast hai, toh $f\to 0^{+}$ (x-axis ke upar se approach karta hai). Jab $x\to -\infty$, $x<0$ aur $e^{-x}\to +\infty$, toh $f\to -\infty$. Shape: $-\infty$ se rise karta hai, origin cross karta hai, $(1,\frac{1}{e})$ par peak karta hai, $x=2$ par inflection se bend over karta hai, phir $y=0$ ki taraf decay karta hai.

Domain: $x-2=0$ at $x=2$: $\mathbb{R}\setminus \{2\}$. Vertical asymptote at $x=2$ (wahan numerator $4-3=1\ne 0$ hai). Intercepts: $f(0)=\frac{-3}{-2}=\frac{3}{2}$. $f=0$ ke liye $x^2-3=0\Rightarrow x=\pm \sqrt{3}\approx \pm 1.732$. Oblique asymptote. Top ki degree ($2$) bottom ($1$) se ek zyada hai → slant line. Polynomial division: $\frac{x^2-3}{x-2}=x+2+\frac{1}{x-2}.$ (Check: $(x+2)(x-2)=x^2-4$, plus remainder $1$ gives $x^2-3$. ✓) Remainder $\frac{1}{x-2}\to 0$ at infinity, toh asymptote $y=x+2$ hai. $x>2$ ke liye uske upar, $x<2$ ke liye neeche. First derivative. Divided form differentiate karo (sabse easy): $f=x+2+(x-2{)}^{-1}$, toh $f^{\prime }(x)=1-(x-2{)}^{-2}=1-\frac{1}{(x-2{)}^2}.$ $f^{\prime }=0$ set karo: $(x-2{)}^2=1\Rightarrow x-2=\pm 1\Rightarrow x=1$ ya $x=3$.

• $f(1)=1+2+\frac{1}{-1}=2$ → $(1,2)$.
• $f(3)=3+2+\frac{1}{1}=6$ → $(3,6)$.

First Derivative Test se classify karo ($f^{\prime }(x)=1-\frac{1}{(x-2{)}^2}$ ka sign). Notice karo ki term $\frac{1}{(x-2{)}^2}$ bada hota hai jab $x$, $2$ ke paas hota hai aur $0$ ki taraf shrink hota hai jab $x$ door hota hai; toh $f^{\prime }>0$ wall se door aur $f^{\prime }<0$ uske paas. Har region mein ek point test karo:

• $x<1$, maano $x=0$: $f^{\prime }=1-\frac{1}{4}=+\frac{3}{4}>0$ → rising.
• $1<x<2$, maano $x=1.5$: $f^{\prime }=1-\frac{1}{0.25}=1-4=-3<0$ → falling.
• $2<x<3$, maano $x=2.5$: $f^{\prime }=1-\frac{1}{0.25}=1-4=-3<0$ → falling.
• $x>3$, maano $x=4$: $f^{\prime }=1-\frac{1}{4}=+\frac{3}{4}>0$ → rising.

Test apply karo:

• $x=1$ par: $f^{\prime }$ changes $+\to -$ ⇒ local maximum at $(1,2)$.
• $x=3$ par: $f^{\prime }$ changes $-\to +$ ⇒ local minimum at $(3,6)$.

Dono branches describe karo (wall $x=2$ curve ko do mein split karti hai):

• Left branch ($x<2$): far left par asymptote $y=x+2$ ke saath $-\infty$ se upar aata hai, local maximum $(1,2)$ tak rise karta hai, phir turn karke $x\to 2^{-}$ par $-\infty$ ki taraf plunge karta hai (denominator $\to 0^{-}$, toh $\frac{1}{x-2}\to -\infty$). Yeh poori branch line $y=x+2$ ke neeche hai kyunki remainder $\frac{1}{x-2}<0$ wahan hai.
• Right branch ($x>2$): $x\to 2^{+}$ par $+\infty$ se descend karta hai (denominator $\to 0^{+}$, toh $\frac{1}{x-2}\to +\infty$), local minimum $(3,6)$ tak pahunchta hai, phir climb karke $x\to +\infty$ par asymptote $y=x+2$ ko upar se hug karta hai (remainder $\frac{1}{x-2}>0$).

Ek interesting baat gaur karne laayak: local minimum $(3,6)$ local maximum $(1,2)$ se upar baithta hai. Yeh bilkul legal hai kyunki dono extrema alag branches par hain jo vertical wall se separated hain — aap ek se doosre tak kabhi continuously travel nahi karte.

General argument. Definition ke hisaab se inflection point woh hota hai jahan $f^{\prime \prime }$ sign change kare. Hume bataya gaya hai ki $f^{\prime \prime }(0)=0$ aur ki $f^{\prime \prime }$, $0$ par sign change karta hai — yahi exactly definition satisfy hoti hai — toh $(0,f(0))$ ek inflection point hai. Kyunki function origin se guzarti hai, $f(0)=0$, toh inflection precisely $(0,0)$ par hai. (Oddness automatically guarantee karti hai $f(0)=0$, kyunki $f(-0)=-f(0)\Rightarrow f(0)=0$.) $f(x)=x^3-3x$ ke saath concrete check:

• Odd? $f(-x)=(-x{)}^3-3(-x)=-x^3+3x=-(x^3-3x)=-f(x)$ ✓.
• Origin se guzar? $f(0)=0^3-3\cdot 0=0$ ✓.
• $f^{\prime \prime }(0)=0$? Pehle $f^{\prime }(x)=3x^2-3$, phir $f^{\prime \prime }(x)=6x$, toh $f^{\prime \prime }(0)=6\cdot 0=0$ ✓.
• $0$ par $f^{\prime \prime }$ ka sign change? $f^{\prime \prime }(-1)=6(-1)=-6<0$ ($0$ ke left mein concave down) aur $f^{\prime \prime }(1)=6(1)=+6>0$ ($0$ ke right mein concave up) — sign $-\to +$ flip hoti hai ✓.

Chaaron hypotheses hold karte hain, toh $(0,0)$, $x^3-3x$ ka inflection point hai, jo parent note ke Worked Example 2 se match karta hai. Yeh Second derivative test aur symmetry ko ek saath tie karta hai.

Maano wall ke perpendicular dono sides $x$ hain, aur parallel side $y$ hai. Fencing used: $2x+y=60\Rightarrow y=60-2x$. Area: $A(x)=xy=x(60-2x)=60x-2x^2,0<x<30.$ (Domain: real side ke liye $x>0$, aur $y=60-2x>0$ force karta hai $x<30$.) First derivative: $A^{\prime }(x)=60-4x$. $=0$ set karo: $x=15$. Second derivative se classify karo: $A^{\prime \prime }(x)=-4<0$ har jagah, toh curve concave down hai ($⌢$), confirm karta hai ki $x=15$ ek maximum hai (yeh Second derivative test hai; Optimization se connect karta hai). Dimensions: $x=15\text{ m}$, $y=60-2(15)=30\text{ m}$. Maximum area $=15\times 30=450{\text{ m}}^2$.