$\begin{array}{l} f(x)=\frac{2-x}{x+1} \\ g(x)=18-3 x \end{array}$ $\newline$ Write $(f \circ g)(x)$ as an expression in terms of $x$ . $\newline$ $(f \circ g)(x)=$

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Domain and range of absolute value functions: equations

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\begin{array}{l} f(x)=\frac{2-x}{x+1} \\ g(x)=18-3 x \end{array}

\newline

Write

(f \circ g)(x)

as an expression in terms of

x

\newline

(f \circ g)(x)=

Understanding $f@g)(x)\:</b> First, let's understand what \$f@g)(x)\ means. The notation \$f@g)(x)\ represents the composition of the functions \$f$ and $g$ , which means we need to apply $g$ first and then apply $f$ to the result of $g$ . In other words, we need to substitute $g(x)$ into $f(x)$ wherever there is an $x$ .
Expressions for $f(x)$ and $g(x)$ : Now let's write down the expressions for $f(x)$ and $g(x)$ to see what we are working with: $\newline$ $f(x) = \frac{2-x}{x+1}$ $\newline$ $g(x) = 18-3x$ $\newline$ We will substitute $g(x)$ into $f(x)$ in place of $x$ .
Substituting $g(x)$ into $f(x)$ : Substitute $g(x)$ into $f(x)$ :
$(f\circ g)(x) = f(g(x)) = f(18-3x) = \frac{2 - (18-3x)}{(18-3x) + 1}$
Simplifying the numerator and denominator: Now, simplify the numerator and the denominator of the fraction: $\newline$ Numerator: $2 - (18 - 3x) = 2 - 18 + 3x = 3x - 16$ $\newline$ Denominator: $(18 - 3x) + 1 = 18 - 3x + 1 = 19 - 3x$ $\newline$ So, $(f@g)(x) = \frac{3x - 16}{19 - 3x}$

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$\begin{array}{l} f(x)=\frac{2-x}{x+1} \\ g(x)=18-3 x \end{array}$ $\newline$ Write $(f \circ g)(x)$ as an expression in terms of $x$ . $\newline$ $(f \circ g)(x)=$