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{"name": "Composite Functions", "extensions": [], "custom_part_types": [], "resources": [], "navigation": {"allowregen": true, "showfrontpage": false, "preventleave": false, "typeendtoleave": false}, "question_groups": [{"pickingStrategy": "all-ordered", "questions": [{"name": "Composite Functions", "tags": [], "metadata": {"description": "Evaluating and finding composite functions.", "licence": "Creative Commons Attribution 4.0 International"}, "statement": "<p>You are given the following functions:</p>\n<p>$f(x) = \\var{a}x+\\var{b}$</p>\n<p>$g(x) = \\var{c}x-\\var{d}$&nbsp;</p>\n<p></p>", "advice": "<p><b>(a)</b></p>\n<p><b>(i)</b></p>\n<p>To find $f(\\var{m}x)$ we just replace&nbsp;$x$ with $\\var{m}x$ in our function definition&nbsp;$f(x) = \\var{a}x+\\var{b}$</p>\n<p>Hence $f(\\var{m}x)&nbsp;= \\var{a}(\\var{m}x)+\\var{b}=\\var{a*m}x+\\var{b}$</p>\n<p></p>\n<p><strong>(ii)</strong></p>\n<p>To find $g(x+\\var{m+1})$ we just replace&nbsp;$x$ with $(x+\\var{m+1})$ in our function definition&nbsp;$g(x) = \\var{c}x-\\var{d}$</p>\n<p>Hence $g(x+\\var{m+1}) = \\var{c}(x+\\var{m+1})-\\var{d}=\\var{c}x+\\var{c*(m+1)}-\\var{d}=\\var{c}x+\\var{c*(m+1)-d}$</p>\n<p></p>\n<p></p>\n<p><strong>(iii)</strong></p>\n<p>$fg(x)=f(g(x))=f(\\var{c}x-\\var{d})=\\var{a}(\\var{c}x-\\var{d})+\\var{b}=\\var{a*c}x-\\var{a*d}+\\var{b}=\\simplify[]{{a*c}x+{b-a*d}}$&nbsp;</p>\n<p></p>\n<p><strong>(iv)</strong></p>\n<p>$gf(x)=g(f(x))=g(\\var{a}x+\\var{b})=\\var{c}(\\var{a}x+\\var{b})-\\var{d}=\\var{a*c}x+\\var{b*c}-\\var{d}=\\simplify[]{{a*c}x+{b*c-d}}$&nbsp;</p>\n<p></p>\n<p><strong>(b)</strong></p>\n<p>Since we found in (ii) that&nbsp;$fg(x)=\\simplify[]{{a*c}x+{b-a*d}}$ we only need to solve the straighfroward equation $\\simplify[]{{a*c}x+{b-a*d}}=\\var{q}$. Hence we obtain $x=\\frac{\\simplify{{q}-{b-a*d}}}{\\var{a*c}}=\\var{sol}$</p>\n<p></p>\n<p>Alternatively, if we hadn't already found the explicit expression for $fg(x)$, we could solve this equation in two steps. Namely, if we can find $y$ such that $f(y)=\\var{q}$, and then find $x$ such that $g(x)=y$, then we must have $f(g(x))=f(y)=\\var{q}$ as required.</p>\n<p>Solving $f(y)= \\var{q}$ gives $\\var{a}y+\\var{b}=&nbsp;\\var{q}.$ Hence $y= \\var{(q-b)/a}$</p>\n<p>Now solving $g(x)=&nbsp; \\var{(q-b)/a}$ gives&nbsp;$\\var{c}x-\\var{d}= \\var{(q-b)/a}$. 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