Question

Help me solve this, Pre-Calculus Question
I only need the last part

Answer 1

`\bf \cfrac{f(a+h)-f(a)}{h}\implies \cfrac{(5a+5h)/(a+h-1)~~-~~(5a)/(a-1)}{h}\impliedby \stackrel{LCD}{(a+h-1)(a-1)} \\\\\\ \cfrac{((a-1)(5a+5h)~~-~~(a+h-1)(5a))/((a+h-1)(a-1))}{h} \\\\\\ \cfrac{(5a^2+5ah-5a-5h~~-~~(5a^2+5ah-5a))/((a+h-1)(a-1))}{h}`


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

Answer 2

first do f(a+h)-f(a)

`(5a+5h)/(a+h-1) - (5a)/(a-1)`

make the denominators the same and simplify:

`((5a+5h)(a-1))/((a+h-1)(a-1)) - (5a(a+h-1))/((a+h-1)(a-1))`

expand:
`(5a^2-5a+5ah-5h-(5a^2+5ah-5a))/((a+h-1)(a-1))`
simplify:
`(-5h)/((a+h-1)(a-1))`

now divide it by h, you have:
`(-5)/((a+h-1)(a-1))`