Question

(f(x)-f(1))/(x-1) if f(x) = ((x+3)/(x+1)), I need to reduce this and I dont know how!

Answer 1

I got tired of trying to write this out sorry:)

Answer 2

\[\frac{\frac{x+3}{x+1}-\frac{1+3}{1+1}}{x-1}=\frac{\frac{x+3}{x+1}-2}{x-1}\]
\[=\frac{x+1}{x+1} \cdot \frac{\frac{x+3}{x+1}-2}{x-1}=\frac{(x+3)-2(x+1)}{(x+1)(x-1)}\]
\[=\frac{-x+1}{(x-1)(x+1)}=\frac{-(x-1)}{(x-1)(x+1)}=\frac{-1}{x+1}\]

Answer 3

and thats of course assuming \[x \neq 1 \]

Answer 4

ok, Thank you so much !! this website is great, I will come here and help people myself more often

Answer 5

and we should also assume
\[x \neq -1\]
since i multiplied by
\[\frac{x+1}{x+1}\]

Answer 6

Yeah, on the second line, but I dont quite understand why?

Answer 7

we have a compound fraction

Answer 8

if i multiplied the top by x+1 it would could clear the denominators on top
but if i multiply by x+1 on top i must do it to the bottom

Answer 9

oh, so you wanted to clean the denominator on top, I see. I though we had to flip the x-1to bring it on top, multiplying the equation by 1/
(x-1)

Answer 10

i don't know what you are saying we could have also combined fractions on top and then change division to multiplication by fliping the second fraction

Answer 11

\[\frac{\frac{x+3}{x+1}-2}{x-1}=\frac{\frac{x+3}{x+1}-2\frac{x+1}{x+1}}{x-1}\]
\[=\frac{x+3-2(x+1)}{x+1} \div (x-1)\]
\[=\frac{-x+1}{x+1} \cdot \frac{1}{x-1}=\frac{-x+1}{(x+1)(x-1)}=\frac{-(x-1)}{(x+1)(x-1)}\]

Answer 12

I agree, but I though the answer was going to be 2 formulas, We initially had to determine the conditions for this function to exist. the x dosnt equal = 1 is good but I though there was something else.

Answer 13

yes
thats the way I though it would be. But just a second

Answer 14

I guess you're right. I had x squares, in my formula, but you just canelled them out. Good so the only condition would be \[x \neq1\]

Answer 15

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