Suppose that the functions $f$ and $g$ are defined as follows. $\newline$ $f(x)=\frac{4}{x+7}\quad g(x)=\frac{5}{x}$ $\newline$ Find $\frac{f}{g}$ . Then, give its domain using an interval or union of intervals. $\newline$ Simplify your answers. $\newline$ $\left(\frac{f}{g}\right)(x)=\prod2$ $\newline$ Domain of $\frac{f}{g}$ : $\newline$ \left[\left(\frac{\square}{\square},\square^{\square},(\square,\square)\right),\left[\square,\square\right],\square\cup\square,(\square,\square]\right),\left[\square,\square\right),\mathcal{O}/,\infty\right),\left[-\infty,\square\right),\left[\times,\(5\right]:\}

View step-by-step help

Home

Math Problems

Algebra 2

Domain and range of absolute value functions: equations

Full solution

Q. Suppose that the functions

f

and

g

are defined as follows.

\newline

f(x)=\frac{4}{x+7}\quad g(x)=\frac{5}{x}

\newline

Find

\frac{f}{g}

. Then, give its domain using an interval or union of intervals.

\newline

Simplify your answers.

\newline

\left(\frac{f}{g}\right)(x)=\prod2

\newline

Domain of

\frac{f}{g}

\newline

\left[\left(\frac{\square}{\square},\square^{\square},(\square,\square)\right),\left[\square,\square\right],\square\cup\square,(\square,\square]\right),\left[\square,\square\right),\mathcal{O}/,\infty\right),\left[-\infty,\square\right),\left[\times,\(5\right]:\}

Dividing Fractions: We have $f(x) = \frac{4}{x+7}$ and $g(x) = \frac{5}{x}$ . To find $(f/g)(x)$ , we divide $f(x)$ by $g(x)$ : $(f/g)(x) = \frac{f(x)}{g(x)} = \frac{\frac{4}{x+7}}{\frac{5}{x}}.$
Multiplying Fractions: To divide the two fractions, we multiply the first fraction by the reciprocal of the second fraction: $\frac{f}{g}\times x = \frac{4}{(x+7)} \times \frac{x}{5} = \frac{4x}{5(x+7)}$ .
Finding Domain: Now we need to find the domain of $(f/g)(x)$ . The domain is the set of all $x$ values for which the function is defined. We need to consider where the denominator is not equal to zero.
Denominator Not Equal to Zero: The denominator of $(f/g)(x)$ is $5(x+7)$ . The function is undefined when the denominator is zero, so we set $5(x+7) \neq 0$ .
Solving for x: Solving for x, we get $x+7 \neq 0$ , which simplifies to $x \neq -7$ . Additionally, since $g(x)$ has a denominator of $x$ , we also have the restriction that $x \neq 0$ .
Domain of $(f/g)(x)$ : Therefore, the domain of $(f/g)(x)$ is all real numbers except $x = -7$ and $x = 0$ . In interval notation, this is $(-\infty, -7) \cup (-7, 0) \cup (0, \infty)$ .