QuestionJune 18, 2025

For f(x)=(x)/(x+1) and g(x)=(7)/(x) find a. (fcirc g)(x) b. the domain of fcirc g a (fcirc g)(x)=square (Simplify your answer.)

For f(x)=(x)/(x+1) and g(x)=(7)/(x) find a. (fcirc g)(x) b. the domain of fcirc g a (fcirc g)(x)=square (Simplify your answer.)
For f(x)=(x)/(x+1) and g(x)=(7)/(x) find a. (fcirc g)(x) b. the domain of fcirc g a (fcirc g)(x)=square (Simplify your answer.)

Solution
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Answer

a. (f \circ g)(x) = \frac{7}{7 + x} ### b. Domain: x \in \mathbb{R}, x \neq 0, x \neq -7 Explanation 1. Find (f \circ g)(x) Substitute g(x) into f(x). So, f(g(x)) = f\left(\frac{7}{x}\right) = \frac{\frac{7}{x}}{\frac{7}{x} + 1}. 2. Simplify the expression Simplify \frac{\frac{7}{x}}{\frac{7}{x} + 1} to get \frac{7}{7 + x} by multiplying numerator and denominator by x. 3. Determine the domain of f \circ g The domain of f \circ g is determined by the restrictions from both f(x) and g(x). For g(x) = \frac{7}{x}, x \neq 0. For f(x) = \frac{x}{x+1}, x \neq -1. Thus, for f(g(x)), x \neq 0 and x \neq -7 (since 7 + x \neq 0).

Explanation

1. Find $(f \circ g)(x)$<br /> Substitute $g(x)$ into $f(x)$. So, $f(g(x)) = f\left(\frac{7}{x}\right) = \frac{\frac{7}{x}}{\frac{7}{x} + 1}$.<br /><br />2. Simplify the expression<br /> Simplify $\frac{\frac{7}{x}}{\frac{7}{x} + 1}$ to get $\frac{7}{7 + x}$ by multiplying numerator and denominator by $x$.<br /><br />3. Determine the domain of $f \circ g$<br /> The domain of $f \circ g$ is determined by the restrictions from both $f(x)$ and $g(x)$. For $g(x) = \frac{7}{x}$, $x \neq 0$. For $f(x) = \frac{x}{x+1}$, $x \neq -1$. Thus, for $f(g(x))$, $x \neq 0$ and $x \neq -7$ (since $7 + x \neq 0$).
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