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Let [tex]f(x)[/tex] be the function [tex]\frac{1}{x+3}[/tex]. Then the quotient [tex]\frac{f(7+h)-f(7)}{h}[/tex] can be simplified to [tex]\frac{-1}{ah+b}[/tex] for:
[tex]a=?[/tex]
[tex]b=?[/tex]

Sagot :

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Step-by-step explanation:

[tex]\frac{f(7+h)-f(7)}{h}\\=\frac{\frac{1}{(7+h)+3}-\frac{1}{7+3}}{h}\\=\frac{\frac{1}{10+h}-\frac{1}{10}}{h}\\=\frac{\frac{10-(10+h)}{10(10+h)}}{h}\\=\frac{-h}{10(10+h)h}}\\=\frac{-1}{10h+100}} \text { for h} \neq 0[/tex]

so a=10, b=100

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