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$\begin{aligned} f(x) &= \dfrac{1}{x-1} \ g(x) &= 5x+8 \end{aligned}$

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Find equations of tangent lines using limits

Full solution

Q.

\begin{aligned} f(x) &= \dfrac{1}{x-1} \ g(x) &= 5x+8 \end{aligned}

Find Composite Function: To find the composite function $f(g(x))$ , we need to substitute $g(x)$ into $f(x)$ where $x$ is. $\newline$ The expression for $f(g(x))$ is $f(5x+8)$ . $\newline$ Let's calculate $f(5x+8)$ by replacing $x$ in $f(x)$ with $5x+8$ . $\newline$ $g(x)$ $0$
Calculate $f(5x+8)$ : Now, simplify the expression inside the parentheses. $f(5x+8) = \frac{1}{5x+8-1}$ $f(5x+8) = \frac{1}{5x+7}$
Find Composite Function: Next, we need to find the composite function $g(f(x))$ . We substitute $f(x)$ into $g(x)$ where $x$ is. $\newline$ The expression for $g(f(x))$ is $g(\frac{1}{x-1})$ . $\newline$ Let's calculate $g(\frac{1}{x-1})$ by replacing $x$ in $g(x)$ with $\frac{1}{x-1}$ . $\newline$ $f(x)$ $0$
Calculate $g\left(\frac{1}{x-1}\right)$ : Now, simplify the expression by distributing the $5$ . $\newline$ $g\left(\frac{1}{x-1}\right) = \frac{5}{x-1} + 8$

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