Respuesta :

first do f(a+h)-f(a)
[tex] \frac{5a+5h}{a+h-1} - \frac{5a}{a-1} [/tex]

make the denominators the same and simplify:
[tex] \frac{(5a+5h)(a-1)}{(a+h-1)(a-1)} - \frac{5a(a+h-1)}{(a+h-1)(a-1)} [/tex]

expand: [tex] \frac{5a^2-5a+5ah-5h-(5a^2+5ah-5a)}{(a+h-1)(a-1)} [/tex]
simplify: [tex] \frac{-5h}{(a+h-1)(a-1)} [/tex]

now divide it by h, you have: [tex] \frac{-5}{(a+h-1)(a-1)} [/tex]
[tex]\bf \cfrac{f(a+h)-f(a)}{h}\implies \cfrac{\frac{5a+5h}{a+h-1}~~-~~\frac{5a}{a-1}}{h}\impliedby \stackrel{LCD}{(a+h-1)(a-1)} \\\\\\ \cfrac{\frac{(a-1)(5a+5h)~~-~~(a+h-1)(5a)}{(a+h-1)(a-1)}}{h} \\\\\\ \cfrac{\frac{5a^2+5ah-5a-5h~~-~~(5a^2+5ah-5a)}{(a+h-1)(a-1)}}{h}[/tex]

[tex]\bf \cfrac{\frac{\underline{5a^2+5ah-5a}-5h~~\underline{-5a^2-5ah+5a}}{(a+h-1)(a-1)}}{h}\implies \cfrac{\frac{-5h}{(a+h-1)(a-1)}}{h} \\\\\\ \cfrac{\frac{-5h}{(a+h-1)(a-1)}}{\frac{h}{1}}\implies \cfrac{-5h}{(a+h-1)(a-1)}\cdot \cfrac{1}{h} \\\\\\ \cfrac{-5\underline{h}}{\underline{h}(a+h-1)(a-1)}\implies \cfrac{-5}{(a+h-1)(a-1)}[/tex]

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