The problem asks to find the domains of functions $f(x) = \sqrt{x + 3}$ and $g(x) = \frac{x}{2}$, as well as the domain of the composition $f \circ g$.

Let’s break it down:

(a) Domain of $f(x) = \sqrt{x + 3}$

For the square root function to be real, the expression inside the square root must be non-negative: $x + 3 \geq 0$ Solving this inequality: $x \geq -3$ Thus, the domain of $f$ is $[-3, \infty)$.

(b) Domain of $g(x) = \frac{x}{2}$

The function $g(x) = \frac{x}{2}$ is defined for all real values of $x$. So, the domain of $g$ is $(-\infty, \infty)$.

(c) Domain of $f \circ g$

The composition $f \circ g$ means $f(g(x)) = \sqrt{g(x) + 3}$.

Since $g(x) = \frac{x}{2}$, we have: $f(g(x)) = \sqrt{\frac{x}{2} + 3}$ For this expression to be defined, the inside of the square root must be non-negative: $\frac{x}{2} + 3 \geq 0$ Solving this inequality: $\frac{x}{2} \geq -3 \quad \Rightarrow \quad x \geq -6$ Thus, the domain of $f \circ g$ is $[-6, \infty)$.

Summary:

• Domain of $f$: $[-3, \infty)$
• Domain of $g$: $(-\infty, \infty)$
• Domain of $f \circ g$: $[-6, \infty)$

Would you like any further details or clarifications?

5 Related Questions:

1. What is the difference between the domain and the range of a function?
2. How do you find the domain of a rational function?
3. Can a function have more than one domain depending on transformations?
4. How would the domain change if $f(x) = \sqrt{2x + 1}$?
5. How does the domain change for compositions of more complex functions?

Tip:

For compositions of functions, always ensure the output of the inner function lies within the domain of the outer function.