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OpenStudy (anonymous):

with f(x)=2x+3, how do I simplify f(x+h)-f(x)/h?

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OpenStudy (anonymous):

\[\frac{f(x+h)-f(x)}{h}=\frac{2(x+h)+3-(2x+3)}{h}=\frac{2x+2h+3-2x-3}{h}=\frac{2h}{h}=2\]

OpenStudy (anonymous):

Yes

OpenStudy (anonymous):

shd101wyy solved the posted question correctly - I assumed this problem is related to limits, for which everything is over h

OpenStudy (anonymous):

because that's how the question was written :)

OpenStudy (anonymous):

great I'm stuck with the same problem, but the f(x)=1/x+1

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OpenStudy (anonymous):

how far did you get ?

OpenStudy (anonymous):

I plug the 1/x+1 in, but the double fractions are killing me...

OpenStudy (anonymous):

is it \[f(x)=\frac{1}{x+1}\] or \[f(x)=\frac{1}{x}+1\]

OpenStudy (anonymous):

the first one

OpenStudy (anonymous):

\[\frac{f(x+h)-f(x)}{h}=\frac{1}{h}*(\frac{1}{x+h+1}-\frac{1}{x+1})=\frac{1}{h}*\frac{x+1-x-h-1}{(x+1)(x+h+1)}\] \[=\frac{1}{h}\frac{-h}{(x+1)(x+h+1)}=-\frac{1}{(x+1)(x+h+1)}\]

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OpenStudy (anonymous):

OHHH I SEE... thanks

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