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aivan3 [116]
4 years ago
11

Pre-cal. Need last part of the answer

Mathematics
1 answer:
erastova [34]4 years ago
8 0
\bf f(x)=\cfrac{3}{x+6}\qquad \qquad \cfrac{f(a+h)-f(a)}{h}\implies \cfrac{\frac{3}{a+h+6}-\frac{3}{a+6}}{h}
\\\\\\
\textit{so our LCD will then be }(a+h+6)(a+6)\textit{ for the numerator}
\\\\\\
\cfrac{\frac{[3(a+6)]~~-~~[3(a+h+6)]}{(a+h+6)(a+6)}}{h}\implies 
\cfrac{\frac{[3a+18]~~-~~[3a+3h+18]}{(a+h+6)(a+6)}}{h}

\bf \cfrac{\frac{\underline{3a+18}~~\underline{-3a}-3h\underline{-18}}{(a+h+6)(a+6)}}{h}\implies 
\cfrac{\frac{-3h}{(a+h+6)(a+6)}}{h}\implies \cfrac{\frac{-3h}{(a+h+6)(a+6)}}{\frac{h}{1}}
\\\\\\
\cfrac{-3h}{(a+h+6)(a+6)}\cdot \cfrac{1}{h}\implies \cfrac{-3\underline{h}}{\underline{h}(a+h+6)(a+6)}
\\\\\\
\cfrac{-3}{(a+h+6)(a+6)}
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