The given problem requires calculating the composite function $(f \circ g)(x)$ and determining its domain $D(f \circ g)$. Here's the solution:

Definitions:

• $f(x) = 3x^2 + 7$
• $g(x) = \sqrt{2x - 1}$

Step 1: Calculate $(f \circ g)(x)$

$(f \circ g)(x) = f(g(x)) = f(\sqrt{2x - 1})$ Substitute $g(x)$ into $f(x)$: $f(\sqrt{2x - 1}) = 3(\sqrt{2x - 1})^2 + 7$ Simplify: $(\sqrt{2x - 1})^2 = 2x - 1$ $f(\sqrt{2x - 1}) = 3(2x - 1) + 7$ $f(\sqrt{2x - 1}) = 6x - 3 + 7 = 6x + 4$

Thus: $(f \circ g)(x) = 6x + 4$

Step 2: Determine the domain $D(f \circ g)$

The domain of $(f \circ g)(x)$ depends on $g(x)$, which must be defined: $g(x) = \sqrt{2x - 1}$ The expression under the square root must be non-negative: $2x - 1 \geq 0 \implies 2x \geq 1 \implies x \geq \frac{1}{2}$

Thus, the domain of $g(x)$ is $x \geq \frac{1}{2}$. Since $f(g(x))$ is valid for all $x \geq \frac{1}{2}$, the domain $D(f \circ g)$ is: $D(f \circ g) = \{x \in \mathbb{R} \mid x \geq \frac{1}{2}\}$

Final Answer:

1. Composite function: $(f \circ g)(x) = 6x + 4$
2. Domain: $D(f \circ g) = [\frac{1}{2}, \infty)$

Would you like further explanation or another example?

Here are 5 related questions for practice:

1. What is the inverse of $g(x) = \sqrt{2x - 1}$?
2. How do you determine the domain of a composite function in general?
3. What is $(g \circ f)(x)$ for the given $f(x)$ and $g(x)$?
4. How does the domain of $g(x)$ affect the domain of $(f \circ g)(x)$?
5. What happens to the composite function when $f(x) = x^2$ and $g(x) = \sqrt{x - 3}$?

Tip: When working with composite functions, always check the domain constraints step-by-step, starting from the inner function.