Let $f(x) = \sqrt{x}$ and $g(x) = x$ be two functions defined in the domain $\mathbf{R}^+ \cup \{0\}$. Find (i) $(f + g)(x)$ (ii) $(f - g)(x)$ (iii) $(fg)(x)$ (iv) $\left(\dfrac{f}{g}\right)(x)$

Let f(x) = √(x) and g(x) = x be two functions defined in the domain mathbfR^+ ∪ \0\. Find (i) (f + g)(x) (ii)

NCERT Exemplar · Q21

f(x) = \sqrt{x}

(i)

(f + g)(x)

(ii)

(f - g)(x)

(iii)

(fg)(x)

(iv)

\left(\dfrac{f}{g}\right)(x)

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Concept understanding — Function Operations

Function Operations: Combining Machines

Think of a function as a machine. You feed it an input (say, a number $x$ ), it does something, and out comes an output $f(x)$ . Now imagine you have two such machines, $f$ and $g$ . Function operations are simply ways to hook these machines together — to add, subtract, multiply, or divide their outputs, or to feed one machine's output into the other.

The core idea is simple: if you can do arithmetic with numbers, you can do arithmetic with functions. The only catch is that both functions must be "ready to work" on the same input at the same time.

The Four Arithmetic Operations

Let $f$ and $g$ be two functions. For any input $x$ that belongs to both their domains (the set of numbers each can accept), we define:

Operation	Notation	What it means
Sum	$(f+g)(x)$	$f(x) + g(x)$
Difference	$(f-g)(x)$	$f(x) - g(x)$
Product	$(f \cdot g)(x)$	$f(x) \cdot g(x)$
Quotient	$\left(\frac{f}{g}\right)(x)$	$\frac{f(x)}{g(x)}$ , provided $g(x) \neq 0$

Note

The domain of the new function is the intersection of the domains of $f$ and $g$ — the numbers both machines can handle. For the quotient, you must also exclude any $x$ where $g(x) = 0$ , because division by zero is undefined.

Example. Let $f(x) = \sqrt{x}$ (domain: $x \geq 0$ ) and $g(x) = x - 1$ (domain: all real numbers). Then:

$(f+g)(x) = \sqrt{x} + x - 1$ , domain: $x \geq 0$ .
$\left(\frac{f}{g}\right)(x) = \frac{\sqrt{x}}{x-1}$ , domain: $x \geq 0$ and $x \neq 1$ .

Composition: Feeding One Machine into Another

This is the most powerful operation. Instead of adding outputs side by side, you take the output of one function and feed it as the input to the other.

(f \circ g)(x) = f\big(g(x)\big)

Read " $f$ composed with $g$ ". You do $g$ first, then $f$ on the result.

Intuition. Suppose $g$ is a machine that converts Celsius to Fahrenheit, and $f$ is a machine that converts Fahrenheit to Kelvin. Then $f \circ g$ converts Celsius directly to Kelvin — one combined machine.

Domain trap. For $f(g(x))$ to make sense, two conditions must hold:

$x$ must be in the domain of $g$ (so $g(x)$ exists).
$g(x)$ must be in the domain of $f$ (so $f$ can accept it).

So the domain of $f \circ g$ is: all $x$ in the domain of $g$ such that $g(x)$ is in the domain of $f$ .

Watch out

Composition is not commutative. $f \circ g$ is almost never the same as $g \circ f$ . For example, if $f(x) = x^2$ and $g(x) = x+1$ , then:

$(f \circ g)(x) = (x+1)^2 = x^2 + 2x + 1$
$(g \circ f)(x) = x^2 + 1$ …

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Function operations combine two functions $f(x)$ and $g(x)$ to form new functions like $(f+g)(x)$ , $(f-g)(x)$ , $(fg)(x)$ , and $\left(\frac{f}{g}\right)(x)$ . The domain of the resulting function is the intersection of the domains of $f$ and $g$ , with an additional restriction for division that the denominator cannot be zero.

$(f+g)(x) = \sqrt{x} + x$ for $x \ge 0$ .
$(f-g)(x) = \sqrt{x} - x$ for $x \ge 0$ .
$(fg)(x) = x^{3/2}$ for $x \ge 0$ .
$\left(\frac{f}{g}\right)(x) = \frac{1}{\sqrt{x}}$ for $x > 0$ .

When we talk about function operations, we are essentially creating new functions by combining existing ones using basic arithmetic operations: addition, subtraction, multiplication, and division. This is very similar to how we combine numbers. If you have two numbers, say $a$ and $b$ , you can find their sum $a+b$ , difference $a-b$ , product $ab$ , and quotient $a/b$ . Functions work in the same way, but instead of operating on single numbers, we operate on their outputs for a given input $x$ .

The crucial aspect to remember is the domain of these new functions. For any operation involving $f(x)$ and $g(x)$ , the new function can only be defined for values of $x$ where both $f(x)$ and $g(x)$ are defined. This means the domain of the resulting function is the intersection of the domains of $f$ and $g$ . For division, there's an additional restriction: the denominator function $g(x)$ cannot be zero.

Let's apply this to the given functions $f(x) = \sqrt{x}$ and $g(x) = x$ .

The problem states that both functions are defined in the domain $\mathbf{R}^+ \cup \{0\}$ , which means $x \ge 0$ .

Determine the common domain of $f(x)$ and $g(x)$ .

The domain of $f(x) = \sqrt{x}$ is $x \ge 0$ .

The domain of $g(x) = x$ is given as $x \ge 0$ .

The intersection of these two domains is $x \ge 0$ . This will be the domain for the sum, difference, and product functions.
(i) Find $(f + g)(x)$ .

The sum of two functions $(f+g)(x)$ is defined as $f(x) + g(x)$ .

Substitute the given expressions for $f(x)$ and $g(x)$ :

$(f+g)(x) = \sqrt{x} + x$

The domain for $(f+g)(x)$ is the common domain of $f$ and $g$ , which is $x \ge 0$ .
(ii) Find $(f - g)(x)$ .

The difference of two functions $(f-g)(x)$ is defined as $f(x) - g(x)$ .

Substitute the given expressions for $f(x)$ and $g(x)$ :

$(f-g)(x) = \sqrt{x} - x$

The domain for $(f-g)(x)$ is the common domain of $f$ and $g$ , which is $x \ge 0$ .
(iii) Find $(fg)(x)$ .

The product of two functions $(fg)(x)$ is defined as $f(x) \cdot g(x)$ .

Substitute the given expressions for $f(x)$ and $g(x)$ :

$(fg)(x) = \sqrt{x} \cdot x$

We can simplify this expression using exponent rules: $\sqrt{x} = x^{1/2}$ and $x = x^1$ .

$(fg)(x) = x^{1/2} \cdot x^1 = x^{(1/2) + 1} = x^{3/2}$

The domain for $(fg)(x)$ is the common domain of $f$ and $g$ , which is $x \ge 0$ .
(iv) Find $\left(\dfrac{f}{g}\right)(x)$ . …

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Q.Let $f(x) = \sqrt{x}$ and $g(x) = x$ be two functions defined in the domain $\mathbf{R}^+ \cup \{0\}$ . Find

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