Chapter 2 · Relations and Functions

2.4 Algebra of Real Functions

Real functions can be added, subtracted, multiplied, divided and scaled point by point — $(f+g)(x)=f(x)+g(x)$ — as long as we keep track of the domain, which shrinks to $D_f\cap D_g$ and, for a quotient, loses the zeros of $g$.

A juice stall at a school fest sells glasses at ₹60 each. Its takings for $x$ glasses are $R(x)=60x$ rupees, and its costs — ₹2000 for the stall and equipment plus ₹35 of fruit per glass — are $C(x)=2000+35x$ rupees. The profit is the difference, $P(x)=R(x)-C(x)=25x-2000$, and the stall breaks even when $P(x)=0$, at $x=80$ glasses. We have just subtracted one function from another, value by value.

New functions are built from old ones like this all the time: total cost is a sum, area is a product of two lengths, average cost is a quotient. In this section we define the sum, difference, product, quotient and scalar multiple of real functions, and — the part students most often get wrong — work out the domain of each.

Sum, Difference and Scalar Multiple

📘 Definition 2.10 — Sum, difference and scalar multiple

Let $f : D_f\to\mathbb{R}$ and $g : D_g\to\mathbb{R}$ be real functions, and let $D=D_f\cap D_g$. For every $x\in D$ define $$(f+g)(x)=f(x)+g(x),\qquad (f-g)(x)=f(x)-g(x).$$ For a real number $\alpha$, the scalar multiple $\alpha f : D_f\to\mathbb{R}$ is $(\alpha f)(x)=\alpha\,f(x)$.

The operations are carried out pointwise: to find $(f+g)(3)$ you find $f(3)$ and $g(3)$ and add the two numbers. The new function is only defined where both ingredients are defined — hence the domain $D_f\cap D_g$, an intersection of sets exactly as in Section 1.3.

✏️ Example 1 — Forming sums, differences and multiples

Let $f(x)=x^2$ and $g(x)=2x+1$, both defined on $\mathbb{R}$. Find $(f+g)(x)$, $(f-g)(x)$, $(3f)(x)$ and $(f-3g)(x)$, and evaluate each at $x=2$.

💡 Show solution▶

Formulas $(f+g)(x)=x^2+2x+1=(x+1)^2$; $(f-g)(x)=x^2-2x-1$; $(3f)(x)=3x^2$; $(f-3g)(x)=x^2-6x-3$. All have domain $\mathbb{R}$.

At x = 2 $f(2)=4$, $g(2)=5$, so $(f+g)(2)=9$, $(f-g)(2)=-1$, $(3f)(2)=12$, $(f-3g)(2)=4-15=-11$.

Check From the formulas: $(2+1)^2=9$ ✓, $4-4-1=-1$ ✓, $3\cdot4=12$ ✓, $4-12-3=-11$ ✓.

✏️ Example 2 — The domain of a sum

Let $f(x)=\sqrt{x-1}$ and $g(x)=\dfrac{1}{x-3}$. Find the domain of $f+g$ and the value $(f+g)(5)$. Is $(f+g)(2)$ defined? Is $(f+g)(0)$?

💡 Show solution▶

Domains $D_f=[1,\infty)$ (need $x-1\ge0$) and $D_g=\mathbb{R}-\{3\}$.

Intersect $D_f\cap D_g=[1,\infty)-\{3\}=[1,3)\cup(3,\infty)$.

Values $(f+g)(5)=\sqrt4+\tfrac12=2.5$. $(f+g)(2)=1+\tfrac{1}{-1}=0$ — defined, since $2$ is in the domain. $(f+g)(0)$ is not defined: $g(0)=-\tfrac13$ exists, but $f(0)=\sqrt{-1}$ is not real.

Product and Quotient

📘 Definition 2.11 — Product and quotient

With $f$, $g$ and $D=D_f\cap D_g$ as before, the product $fg : D\to\mathbb{R}$ and the quotient $\dfrac fg$ are $$(fg)(x)=f(x)\,g(x)\quad(x\in D),$$ $$\left(\frac fg\right)(x)=\frac{f(x)}{g(x)}\quad(x\in D,\ g(x)\ne0).$$

📐 Theorem 2.5 — Domains of combined functions

For real functions $f$ and $g$ with domains $D_f$ and $D_g$, and any real $\alpha$:

$f+g$, $f-g$ and $fg$ have domain $D_f\cap D_g$;
$\alpha f$ has domain $D_f$;
$\dfrac fg$ has domain $\big(D_f\cap D_g\big)-\{x : g(x)=0\}$.

Proof. The number $f(x)\pm g(x)$ or $f(x)g(x)$ exists exactly when both $f(x)$ and $g(x)$ exist, i.e. when $x\in D_f$ and $x\in D_g$. The number $\alpha f(x)$ needs only $f(x)$. The quotient needs both values and, in addition, a non-zero denominator, so the zeros of $g$ are removed. $\blacksquare$

✏️ Example 3 — Products and quotients of polynomials

Let $f(x)=x+1$ and $g(x)=x-2$. Find $(fg)(x)$ and $\left(\dfrac fg\right)(x)$ with their domains, and evaluate both at $x=4$ and $x=2$.

💡 Show solution▶

Product $(fg)(x)=(x+1)(x-2)=x^2-x-2$, domain $\mathbb{R}$. $(fg)(4)=5\cdot2=10$; $(fg)(2)=3\cdot0=0$.

Quotient $\left(\dfrac fg\right)(x)=\dfrac{x+1}{x-2}$; $g(x)=0$ at $x=2$, so the domain is $\mathbb{R}-\{2\}$.

Values $\left(\dfrac fg\right)(4)=\dfrac52=2.5$; $\left(\dfrac fg\right)(2)$ is not defined.

✏️ Example 4 — A quotient that simplifies

Let $f(x)=x^2-4$ and $g(x)=x-2$. Simplify $\dfrac fg$, give its domain, and describe its graph.

💡 Show solution▶

Domain first $g(x)=0$ at $x=2$, so the domain of $\dfrac fg$ is $\mathbb{R}-\{2\}$. Decide this before simplifying.

Simplify For $x\ne2$: $\dfrac{x^2-4}{x-2}=\dfrac{(x-2)(x+2)}{x-2}=x+2$.

Graph The line $y=x+2$ with a single point missing: a hollow dot at $(2,4)$. So $\dfrac fg$ and the function $x+2$ are different functions — they have different domains.

The purple curve is $\dfrac{f}{g}$. Wherever $g(x)=0$ the quotient is undefined: if $f(x)$ is zero there too, the graph just loses a point (a hollow dot); otherwise the values shoot off to $\pm$ very large numbers along a dashed vertical line. The probe slows down near each excluded $x$ — watch the readout.

Quotient:

$x$—

$g(x)$—

$(f/g)(x)$—

domain of $f/g$—

✏️ Example 5 — Profit and break-even

For the fest stall in the introduction, $R(x)=60x$ and $C(x)=2000+35x$ (in rupees) for $x$ glasses. (a) Find the profit function $P=R-C$. (b) Find the break-even number of glasses. (c) Find the profit on $120$ glasses. (d) Find the average cost per glass $A=\dfrac Cx$ for $x=100$, and its domain.

💡 Show solution▶

(a) $P(x)=60x-(2000+35x)=25x-2000$ rupees.

(b) $P(x)=0$ gives $x=80$ glasses: below $80$ the stall makes a loss, above $80$ a profit.

(d) $A(x)=\dfrac{2000+35x}{x}=\dfrac{2000}{x}+35$; $A(100)=20+35=55$, i.e. ₹55 per glass. The quotient needs $x\ne0$, and in context $x$ is a positive whole number, so the domain is $\mathbb{N}$.

✏️ Example 6 — Combining standard functions

Let $f(x)=|x|$ and $g(x)=x$. Find $(f+g)(x)$, $(fg)(x)$ and $\left(\dfrac fg\right)(x)$, writing each without the modulus sign, and give the domain of each.

💡 Show solution▶

Split Use $|x|=x$ for $x\ge0$ and $|x|=-x$ for $x\lt0$.

Sum $(f+g)(x)=\begin{cases}2x, & x\ge0\\ 0, & x\lt0\end{cases}$ with domain $\mathbb{R}$ — flat on the left, slope $2$ on the right.

Product $(fg)(x)=x|x|=\begin{cases}x^2, & x\ge0\\ -x^2, & x\lt0\end{cases}$ with domain $\mathbb{R}$.

Quotient $\left(\dfrac fg\right)(x)=\dfrac{|x|}{x}=\begin{cases}1, & x\gt0\\ -1, & x\lt0\end{cases}$ with domain $\mathbb{R}-\{0\}$. It agrees with $\operatorname{sgn}x$ except at $0$, where $\operatorname{sgn}0=0$ but $\dfrac fg$ is undefined.

⚠️ Common mistakes in the algebra of functions

1. Forgetting to intersect domains. $\sqrt{x}+\dfrac1{x-1}$ is not defined at $x=-4$ (first term) or at $x=1$ (second term).

2. Cancelling away an excluded point. Find the domain of $\dfrac fg$ from the original $g$, before simplifying (Example 4).

3. $fg$ is a product. $(fg)(x)$ means $f(x)\times g(x)$, never "$f$ applied to $g(x)$". With $f(x)=x+1$, $g(x)=x^2$: $(fg)(2)=3\times4=12$.

4. Brackets in a difference. $(f-g)(x)=f(x)-g(x)$ subtracts all of $g(x)$: $x^2-(2x+1)=x^2-2x-1$, not $x^2-2x+1$.

📘 In the CBSE syllabus — school assessment only (not in the year-end board exam)

Composition of functions is taught and tested in school, but not in the year-end board exam; JEE Main and JEE Advanced both list it, and you will use it constantly in Class 12 (inverse functions, the chain rule).

Composition. Mistake 3 above warned that $fg$ is a product. "Apply $f$, then apply $g$ to the result" is a different way of combining two functions: the composite $g\circ f$ ("$g$ of $f$"), defined by $$(g\circ f)(x)=g\big(f(x)\big).$$ Its domain is the set of $x$ in the domain of $f$ for which $f(x)$ lies in the domain of $g$ — the output of the first machine must be an allowed input of the second. The function written on the right acts first, and order matters: in general $g\circ f\ne f\circ g$.

Worked example 1. Let $f(x)=x+1$ and $g(x)=x^2$. Find $(g\circ f)(x)$ and $(f\circ g)(x)$, and compare them at $x=2$.
Solution. $(g\circ f)(x)=g(x+1)=(x+1)^2$, while $(f\circ g)(x)=f(x^2)=x^2+1$. At $x=2$: $(g\circ f)(2)=9$ but $(f\circ g)(2)=5$, so $g\circ f\ne f\circ g$. Both have domain $\mathbb{R}$.

Worked example 2. Let $f(x)=x+3$ and $g(x)=\sqrt x$. Find $g\circ f$ and $f\circ g$ with their domains.
Solution. $(g\circ f)(x)=\sqrt{x+3}$: we need $f(x)=x+3\ge0$, so the domain is $[-3,\infty)$. $(f\circ g)(x)=\sqrt x+3$: now $x$ itself must be $\ge0$, so the domain is $[0,\infty)$. Check $x=-2$: $g(f(-2))=g(1)=1$ is defined, but $g(-2)$ is not, so $-2$ is in the first domain only ✓.

🚀 Beyond the board syllabus (JEE Main · JEE Advanced)

In JEE Main and JEE Advanced 2026; the CBSE year-end exam asks only for domains of simple sums, products and quotients. Both entrance exams ask for domains of expressions that combine several roots, quotients and standard functions.

Method. Treat the expression as a sum, product or quotient of simple pieces, write the domain of each piece, intersect them (Theorem 2.5), and remove every zero of every denominator.

Worked example. Find the domain of $h(x)=\sqrt{x-2}+\dfrac{1}{\sqrt{5-x}}+\dfrac{1}{x-4}$.
Solution. $\sqrt{x-2}$ needs $x\ge2$. $\dfrac1{\sqrt{5-x}}$ needs $5-x\gt0$ (strictly, as it is in a denominator), i.e. $x\lt5$. $\dfrac1{x-4}$ needs $x\ne4$. Intersecting: $[2,5)-\{4\}=[2,4)\cup(4,5)$.

Exercises

Combining two functions. In parts 1–6, $f(x)=x+1$ and $g(x)=x^2$.

1 Find $(f+g)(x)$.
2 Find $(f-g)(2)$.
3 Find $(fg)(-1)$.
4 Find $\left(\dfrac fg\right)(x)$ and its domain.
5 Find $(3f-2g)(1)$.
6 Find $\left(\dfrac gf\right)(x)$ and its domain.
7 $f(x)=3x-2$ and $g(x)=-3x+5$. Show that $f+g$ is a constant function.

💡 Show solutions▶

1 $x^2+x+1$, domain $\mathbb{R}$.

2 $f(2)-g(2)=3-4=-1$.

3 $f(-1)g(-1)=0\cdot1=0$.

4 $\dfrac{x+1}{x^2}$; $g(x)=0$ only at $0$, so the domain is $\mathbb{R}-\{0\}$.

5 $3f(1)-2g(1)=6-2=4$.

6 $\dfrac{x^2}{x+1}$, domain $\mathbb{R}-\{-1\}$.

7 $(f+g)(x)=3x-2-3x+5=3$ for every $x$: a constant function with range $\{3\}$.

Domains. Find the domain of the stated combination.

1 $f+g$, where $f(x)=\sqrt x$ and $g(x)=\sqrt{4-x}$.
2 $f+g$, where $f(x)=\sqrt{x-1}$ and $g(x)=\dfrac1{x-2}$.
3 $\dfrac fg$, where $f(x)=x$ and $g(x)=x^2-9$.
4 $\dfrac fg$, where $f(x)=\sqrt{x+2}$ and $g(x)=x-1$.
5 $fg$, where $f(x)=\dfrac1x$ and $g(x)=\dfrac1{x-1}$.
6 $\dfrac fg$, where $f(x)=x^2-1$ and $g(x)=x+1$. Simplify it and find its value at $x=3$.
7 $\dfrac fg$, where $f(x)=x+3$ and $g(x)=|x|-2$.

Check: the domain of each part on the number line

Part:

domain—

excluded or cut off by—

💡 Show solutions▶

1 $D_f=[0,\infty)$, $D_g=(-\infty,4]$; domain $[0,4]$.

2 $[1,\infty)\cap(\mathbb{R}-\{2\})=[1,2)\cup(2,\infty)$.

3 $x^2-9=0$ at $x=\pm3$: domain $\mathbb{R}-\{-3,3\}$.

4 $D_f=[-2,\infty)$; remove $x=1$: $[-2,1)\cup(1,\infty)$.

5 $(\mathbb{R}-\{0\})\cap(\mathbb{R}-\{1\})=\mathbb{R}-\{0,1\}$.

6 Domain $\mathbb{R}-\{-1\}$; for $x\ne-1$, $\dfrac{(x-1)(x+1)}{x+1}=x-1$; value at $3$: $2$.

7 $|x|-2=0$ at $x=\pm2$: domain $\mathbb{R}-\{-2,2\}$.

III

Combining standard functions.

1 $f(x)=|x|$, $g(x)=x$. Write $(f-g)(x)$ without the modulus sign.
2 With $f$, $g$ as in part 1, find $(f-g)(-3)$ and $(f-g)(3)$.
3 $h(x)=x-[x]$. Find $h(2.3)$, $h(-2.3)$ and $h(5)$, and the range of $h$.
4 $f(x)=2x$, $g(x)=x^2$. Find $\dfrac fg$ and $\dfrac gf$ with their domains.
5 $f(x)=\operatorname{sgn}x$, $g(x)=|x|$. Identify the function $fg$.
6 $f(x)=x^2$. Describe the graphs of $2f$, $\tfrac12f$ and $(-1)f$ compared with the graph of $f$.

💡 Show solutions▶

1 $(f-g)(x)=\begin{cases}0, & x\ge0\\ -2x, & x\lt0.\end{cases}$

2 $(f-g)(-3)=3-(-3)=6$; $(f-g)(3)=0$.

3 $h(2.3)=2.3-2=0.3$; $h(-2.3)=-2.3-(-3)=0.7$; $h(5)=0$. Since $[x]\le x\lt[x]+1$, $0\le h(x)\lt1$, and every value in $[0,1)$ occurs (e.g. $h(t)=t$ for $0\le t\lt1$): range $[0,1)$.

4 $\dfrac fg=\dfrac{2x}{x^2}=\dfrac2x$ with domain $\mathbb{R}-\{0\}$; $\dfrac gf=\dfrac{x^2}{2x}=\dfrac x2$ with domain $\mathbb{R}-\{0\}$ — the zero of $f$ is removed even though $\dfrac x2$ makes sense at $0$.

5 $\operatorname{sgn}x\cdot|x|=x$ for every $x$ (Section 2.3, Example 4), so $fg$ is the identity function.

6 $2f$: the parabola stretched vertically by $2$ (narrower). $\tfrac12f$: squashed vertically (wider). $(-1)f$: reflected in the $x$-axis, opening downwards, range $(-\infty,0]$.

Applications.

1 Business. A start-up selling notebooks has revenue $R(x)=45x$ and cost $C(x)=1500+20x$ rupees for $x$ notebooks. Find the profit function, the break-even point, and the profit on $100$ notebooks.
2 Area. A rectangle has sides $f(x)=x+2$ and $g(x)=x-1$ metres, where $x\gt1$. Express its area as a function and evaluate it at $x=4$.
3 Average cost. Printing $x$ copies of a school magazine costs $C(x)=500+10x$ rupees. Find the average cost function $A(x)=\dfrac{C(x)}{x}$, its domain in this context, and $A(50)$.
4 Motion. Two cars leave a toll plaza together on the same road; after $t$ hours car P has gone $d(t)=60t$ km and car Q $e(t)=45t$ km. Find the gap $(d-e)(t)$ and when it is $30$ km.
5 Electricity. A monthly bill is $B(x)=100+6x$ rupees for $x$ units. Flat 1 uses $x$ units and Flat 2 uses twice as many. Write the combined bill $T(x)=B(x)+B(2x)$ as a formula and evaluate it for $x=50$.

💡 Show solutions▶

1 $P(x)=25x-1500$ rupees; break-even at $x=60$ notebooks; $P(100)=1000$, a profit of ₹1000.

2 $(fg)(x)=(x+2)(x-1)=x^2+x-2$ square metres; at $x=4$: $6\times3=18\ \text{m}^2$.

3 $A(x)=\dfrac{500}{x}+10$ rupees per copy; domain: positive whole numbers $\mathbb{N}$ (the formula alone would allow any $x\ne0$); $A(50)=20$, i.e. ₹20 per copy.

4 $(d-e)(t)=15t$ km; $15t=30$ gives $t=2$ hours.

5 $T(x)=(100+6x)+(100+12x)=200+18x$ rupees; $T(50)=200+900=1100$, i.e. ₹1100.

📘 School-assessment practice (composition — see the 📘 box; not in the year-end board exam). In parts 1–2, $f(x)=2x-1$ and $g(x)=x^2+2$.

1 Find $(g\circ f)(x)$ and $(f\circ g)(x)$.
2 Find $(g\circ f)(1)$ and $(f\circ g)(1)$.
3 For $f(x)=\sqrt x$ and $g(x)=1-x$, find $g\circ f$ and $f\circ g$ with their domains.
4 For $f(x)=x^2$, find $(f\circ f)(x)$ and $(f\circ f)(-2)$.
5 If $f(x)=3x+2$, find a function $g$ with $(f\circ g)(x)=x$ for every real $x$, and check that $(g\circ f)(x)=x$ too.

💡 Show solutions▶

1 $(g\circ f)(x)=(2x-1)^2+2=4x^2-4x+3$; $(f\circ g)(x)=2(x^2+2)-1=2x^2+3$.

2 $(g\circ f)(1)=g(1)=3$; $(f\circ g)(1)=f(3)=5$.

3 $(g\circ f)(x)=1-\sqrt x$, domain $[0,\infty)$; $(f\circ g)(x)=\sqrt{1-x}$, which needs $1-x\ge0$: domain $(-\infty,1]$.

4 $(f\circ f)(x)=(x^2)^2=x^4$; $(f\circ f)(-2)=16$.

5 We need $3g(x)+2=x$, so $g(x)=\dfrac{x-2}{3}$. Then $(g\circ f)(x)=\dfrac{(3x+2)-2}{3}=x$ ✓ — $g$ undoes $f$ (an inverse function, Class 12).

🎯 Exam corner Board · 1 markBoard · 2 marksBoard · 3 marksJEE MainJEE Advanced

How this is examined: Board papers give two functions and ask for $f\pm g$, $fg$, $f/g$ — as formulas, as values at a point, and with domains (1–3 marks). JEE combines the standard functions ($|x|$, $[x]$) and asks for values, domains and identities.

E1. Board · 1 mark If $f(x)=x^2$ and $g(x)=2x+1$, then $(f-g)(3)$ equals (A) $2$ (B) $16$ (C) $-2$ (D) $4$

💡 Show solution▶

Idea $f(3)=9$, $g(3)=7$, and $(f-g)(3)=9-7$.

Trap (B) is $(f+g)(3)$; (D) comes from $9-(6-1)$, a sign slip in the brackets.

Answer (A).

E2. Board · 2 marks Let $f(x)=\sqrt{x-1}$ and $g(x)=x-3$. Find $(f+g)(x)$, $(fg)(x)$ and $\left(\dfrac fg\right)(x)$, stating the domain of each.

💡 Show solution▶

Domains $D_f=[1,\infty)$, $D_g=\mathbb{R}$, so $D_f\cap D_g=[1,\infty)$. (1 mark)

Answer $(f+g)(x)=\sqrt{x-1}+x-3$ and $(fg)(x)=(x-3)\sqrt{x-1}$, each on $[1,\infty)$; $\left(\dfrac fg\right)(x)=\dfrac{\sqrt{x-1}}{x-3}$ on $[1,3)\cup(3,\infty)$. (1 mark)

E3. Board · 3 marks Let $f(x)=x^2-9$ and $g(x)=x+3$. Find (i) $(f+g)(x)$ (ii) $(fg)(1)$ (iii) $\left(\dfrac fg\right)(x)$ in simplest form, with its domain.

💡 Show solution▶

(i) $x^2-9+x+3=x^2+x-6$. (1 mark)

(ii) $f(1)g(1)=(-8)(4)=-32$. (1 mark)

Answer (iii) $g(x)=0$ at $x=-3$, so the domain is $\mathbb{R}-\{-3\}$; there $\dfrac{(x-3)(x+3)}{x+3}=x-3$. So $\left(\dfrac fg\right)(x)=x-3$, $x\ne-3$. (1 mark)

E4. JEE Main If $f(x)=|x|$ and $g(x)=[x]$, then $(f-g)(-1.5)$ is (A) $3.5$ (B) $-0.5$ (C) $0.5$ (D) $2.5$

💡 Show solution▶

Idea $|-1.5|=1.5$ and $[-1.5]=-2$, so $(f-g)(-1.5)=1.5-(-2)$.

Trap (D) uses $[-1.5]=-1$; (B) uses $|-1.5|=-1.5$ together with $[-1.5]=-1$.

Answer (A) $3.5$.

E5. JEE Main Numerical value. Find the number of integers in the domain of $h(x)=\dfrac{\sqrt{x-1}\,\sqrt{7-x}}{x-4}$.

💡 Show solution▶

Pieces $\sqrt{x-1}$: $x\ge1$. $\sqrt{7-x}$: $x\le7$. Denominator: $x\ne4$. Domain $[1,7]-\{4\}$ (the 🚀 box method).

Answer Integers $1,2,3,5,6,7$: $6$.

E6. JEE Advanced One or more options correct. Let $f(x)=x+|x|$ and $g(x)=x-|x|$ for $x\in\mathbb{R}$. Then (A) $(fg)(x)=0$ for every real $x$ (B) $(f+g)(x)=2x$ for every real $x$ (C) the domain of $\dfrac fg$ is $(-\infty,0]$ (D) the range of $g$ is $(-\infty,0)$

💡 Show solution▶

Pieces For $x\ge0$: $f(x)=2x$, $g(x)=0$. For $x\lt0$: $f(x)=0$, $g(x)=2x$.

(A), (B) In each case one factor is $0$, so $fg=0$: true. $f+g=2x$ directly: true.

(C), (D) $g(x)=0$ for all $x\ge0$, so $\dfrac fg$ is defined only for $x\lt0$: domain $(-\infty,0)$, not $(-\infty,0]$ — false. The range of $g$ is $(-\infty,0]$ (it includes $0$) — false.

Answer (A), (B).

🔬 Research–Teaching NexusMini project · about 50 minutes

Research question. For a real small venture — a school-fest stall, a tuition group, a home bakery order — what are the revenue, cost and profit functions, and how many sales are needed to break even?

Why it matters. Break-even analysis is how every business, from a street vendor to an airline, decides whether a plan is worth starting. It is nothing more than the algebra of functions: $P=R-C$, average cost $C/x$, and the point where $P$ changes sign.

What to do.

Choose a venture and collect real prices (ask a shopkeeper, check online rates): the fixed costs, the cost per item, and a realistic selling price.
Write $R(x)$, $C(x)$ and $P(x)=(R-C)(x)$, and state a sensible domain for $x$.
Find the break-even point and the profit at two realistic sales levels. Tabulate the average cost $\dfrac{C(x)}{x}$ for $x=10,50,100,200$.
Change one price (e.g. raise the selling price by ₹5) and recompute the break-even point.

Tools: A spreadsheet with columns $x$, $R$, $C$, $P$, $C/x$; a phone graphing app

Hand in: The three functions, a graph of $R$ and $C$ on the same axes with the break-even point marked, and a recommendation in two sentences.

Going further. Why does the average cost $\dfrac{C(x)}{x}$ keep falling as $x$ grows, but never below the cost per item? Explain using $\dfrac{C(x)}{x}=\dfrac{\text{fixed cost}}{x}+\text{cost per item}$.

Instructor note: real cost data are rarely exactly linear (bulk discounts, extra staff); encourage students to state the assumptions behind their straight-line model and the range of $x$ for which it is believable.

🧪 Self-check quiz

0/5 correct

Q1. If $f(x)=x+2$ and $g(x)=3x$, then $(f+g)(1)$ is

Q2. If $f(x)=x$ and $g(x)=x-5$, the domain of $\dfrac fg$ is

Q3. If $f(x)=x^2$, then $(2f)(x)$ is

Q4. If $f(x)=\sqrt x$ and $g(x)=x+1$, the domain of $f+g$ is

Q5. If $f(x)=x-1$ and $g(x)=x+1$, then $(fg)(x)$ is

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