Q.Let $f(x) = \sqrt{x}$ and $g(x) = x$ be two functions defined over the set of non-negative real numbers. Find $(f + g)(x)$, $(f - g)(x)$, $(fg)(x)$ and $\left(\dfrac{f}{g}\right)(x)$.

The key idea is that function operations are performed pointwise — each new function is defined by applying the operation to the outputs of $f$ and $g$ at the same $x$, using the same domain where both original functions are defined.

Since $f(x) = \sqrt{x}$ and $g(x) = x$ are both defined for $x \geq 0$, the domain for all combined functions is $x \geq 0$, except division where $g(x) \neq 0$ (so $x > 0$).

1. $(f + g)(x) = \sqrt{x} + x$, for $x \geq 0$.
2. $(f - g)(x) = \sqrt{x} - x$, for $x \geq 0$.
3. $(fg)(x) = \sqrt{x} \cdot x = x^{3/2}$, for $x \geq 0$. …

The core idea here is that when we have two functions defined on the same domain, we can combine them by performing arithmetic operations on their outputs at each input. This is called pointwise operation — for each $x$, we compute $f(x)$ and $g(x)$ separately, then add, subtract, multiply, or divide those numbers.

Since both $f$ and $g$ are defined for $x \geq 0$, the combined functions will also be defined there — except for division, where we must exclude points where $g(x)=0$.

Let’s work through each operation step by step.

1. Addition: $(f+g)(x)$

   By definition, $(f+g)(x) = f(x) + g(x)$.

   Here $f(x) = \sqrt{x}$ and $g(x) = x$, so:

The domain remains $x \geq 0$, since both terms are defined there.

2. Subtraction: $(f-g)(x)$

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

   Domain: $x \geq 0$.

3. Multiplication: $(fg)(x)$

   $(fg)(x) = f(x) \cdot g(x) = \sqrt{x} \cdot x$.

   Recall that $\sqrt{x} \cdot x = x^{1/2} \cdot x^1 = x^{3/2}$.

   So $(fg)(x) = x^{3/2}$, with domain $x \geq 0$.

4. Division: $\left(\frac{f}{g}\right)(x)$

   $\left(\frac{f}{g}\right)(x) = \frac{f(x)}{g(x)} = \frac{\sqrt{x}}{x}$, provided $g(x) \neq 0$.

   Since $g(x)=x$, we must exclude $x=0$. …