Intermediate Algebra: Connecting Concepts through Application

$\text{a) } f(g(x))=\dfrac{1}{6}x+\dfrac{5}{6} \\\\\text{b) } f(42)g(42)=\dfrac{1,275}{4} \\\\\text{c) } f(x)+g(x)=\dfrac{11}{12}x+\dfrac{13}{12}$
$\bf{\text{Solution Outline:}}$ Substitute the given functions, \begin{array}{l}\require{cancel} f(x)= \dfrac{2}{3}x+\dfrac{1}{3} \\g(x)= \dfrac{1}{4}x+\dfrac{3}{4} ,\end{array} into the required operations. $\bf{\text{Solution Details:}}$ a) \begin{array}{l}\require{cancel} f(g(x))=f\left( \dfrac{1}{4}x+\dfrac{3}{4} \right) \\\\ f(g(x))=\dfrac{2}{3}\left( \dfrac{1}{4}x+\dfrac{3}{4} \right)+\dfrac{1}{3} \\\\ f(g(x))=\dfrac{2}{12}x+\dfrac{6}{12} +\dfrac{1}{3} \\\\ f(g(x))=\dfrac{1}{6}x+\dfrac{1}{2} +\dfrac{1}{3} \\\\ f(g(x))=\dfrac{1}{6}x+\dfrac{3}{6} +\dfrac{2}{6} \\\\ f(g(x))=\dfrac{1}{6}x+\dfrac{5}{6} \end{array} b) \begin{array}{l}\require{cancel} f(x)g(x)=\left( \dfrac{2}{3}x+\dfrac{1}{3} \right) \left( \dfrac{1}{4}x+\dfrac{3}{4} \right) \\\\ f(42)g(42)=\left( \dfrac{2}{3}(42)+\dfrac{1}{3} \right) \left( \dfrac{1}{4}(42)+\dfrac{3}{4} \right) \\\\ f(42)g(42)=\left( \dfrac{84}{3}+\dfrac{1}{3} \right) \left( \dfrac{42}{4}+\dfrac{3}{4} \right) \\\\ f(42)g(42)=\left( \dfrac{85}{3} \right) \left( \dfrac{45}{4} \right) \\\\ f(42)g(42)=\left( \dfrac{85}{\cancel{3}} \right) \left( \dfrac{\cancel{3}(15)}{4} \right) \\\\ f(42)g(42)=\dfrac{1,275}{4} \end{array} c) \begin{array}{l}\require{cancel} f(x)+g(x)=\left( \dfrac{2}{3}x+\dfrac{1}{3} \right) + \left( \dfrac{1}{4}x+\dfrac{3}{4} \right) \\\\ f(x)+g(x)=\left( \dfrac{2}{3}x+\dfrac{1}{4}x \right) + \left(\dfrac{3}{4}+\dfrac{1}{3} \right) \\\\ f(x)+g(x)=\left( \dfrac{8}{12}x+\dfrac{3}{12}x \right) + \left(\dfrac{9}{12}+\dfrac{4}{12} \right) \\\\ f(x)+g(x)=\dfrac{11}{12}x+\dfrac{13}{12} \end{array} Therefore, \begin{array}{l}\require{cancel} \text{a) } f(g(x))=\dfrac{1}{6}x+\dfrac{5}{6} \\\\\text{b) } f(42)g(42)=\dfrac{1,275}{4} \\\\\text{c) } f(x)+g(x)=\dfrac{11}{12}x+\dfrac{13}{12} .\end{array}