QuestionSuppose that the functions $f$ and $g$ are defined as follows. $$\begin{array}{l} f(x)=\sqrt{x+4} \\ g(x)=\frac{1}{3 x^{2}+1} \end{array}$$
Find $f \cdot g$ and $f-g$. Then, give their domains using interval notation. $$(f \cdot g)(x)=$$ $\square$
Domain of $f \cdot g$ : $\square$ $$(f-g)(x)=$$ $\square$
Domain of $f-g$ : $\square$

Studdy Solution
Determine the domain of $f - g$:
- For $f(x) = \sqrt{x+4}$, the condition is $x \geq -4$. - For $g(x) = \frac{1}{3x^2 + 1}$, there are no additional restrictions as the denominator is always positive.
Thus, the domain of $f - g$ is also $x \geq -4$.
Domain of $f - g$: $[-4, \infty)$
The expressions and domains are:
$(f \cdot g)(x) = \sqrt{x+4} \cdot \frac{1}{3x^2 + 1}$
Domain of $f \cdot g$: $[-4, \infty)$
$(f - g)(x) = \sqrt{x+4} - \frac{1}{3x^2 + 1}$
Domain of $f - g$: $[-4, \infty)$