The image provided shows the problem asking to find the formula for the new function formed by the composition $(f \circ g \circ h)(x)$, where the given functions are:

$f(x) = \sqrt{x+3}, \quad g(x) = \frac{6}{x}, \quad h(x) = x^2.$

Let's break down the composition step-by-step:

1. Compute $(g \circ h)(x)$:

We need to substitute $h(x) = x^2$ into $g(x) = \frac{6}{x}$:

$g(h(x)) = g(x^2) = \frac{6}{x^2}.$

2. Now, compute $(f \circ g \circ h)(x)$:

Substitute $g(h(x)) = \frac{6}{x^2}$ into $f(x) = \sqrt{x + 3}$:

$f(g(h(x))) = f\left(\frac{6}{x^2}\right) = \sqrt{\frac{6}{x^2} + 3}.$

Thus, the composed function is:

$(f \circ g \circ h)(x) = \sqrt{\frac{6}{x^2} + 3}.$

Domain:

To determine the domain, we need to consider the restrictions of each function:

• $h(x) = x^2$ is defined for all real $x$, $(-\infty, \infty)$.
• $g(x) = \frac{6}{x}$ has a restriction that $x \neq 0$.
• $f(x) = \sqrt{x+3}$ requires that $x + 3 \geq 0$, or $x \geq -3$.

Thus, after combining these conditions, the domain will be restricted by $h(x) \neq 0$, meaning $x \neq 0$. However, since $h(x) = x^2$, $h(x)$ is always non-negative, and the square root function restricts it further.

Therefore, the domain is:

$(0, \infty).$

Let me know if you would like further clarifications or detailed explanations!

Follow-up questions:

1. How is the domain determined in function compositions?
2. What are the possible cases where the domain of a composed function could be restricted further?
3. How do you handle square roots in function composition?
4. Can you explain how the square in $h(x) = x^2$ affects the domain of the composition?
5. Why does $g(x) = \frac{6}{x}$ cause issues when $x = 0$?

Tip:

Always evaluate the inner functions first in function composition and ensure the domains align before simplifying the entire expression.