Question

Evaluate the difference quotient f(x) = x + 3 x + 1 , f(x) − f(1) x − 1

Answer 1

maybe
\[\frac{f(x)-f(1)}{x-1}\]?

Answer 2

i dont know ive been looking for two days now and im still clueless

Answer 3

i mean is that what you have to compute? it is not exactly clear from what you wrote
if that is the what you need, we can do it no problem

Answer 4

dang just realized i had the wrong thing wrote
F(x) = (x+3)/(x+1) , (f(x) - f(10)) / (x-1)

Answer 5

maybe
\[\frac{f(x)-f(1)}{x-1}\]?

Answer 6

good lord i give up at typing, its to late
F(x) = (x+3)/(x+1) , (f(x) - f(1)) / (x-1)

Answer 7

ok ready?

Answer 8

i think so

Answer 9

first we need \(f(1)\) which is easy enough, since
\[f(x)=\frac{x+3}{x+1}\] we know
\[f(1)=\frac{1+3}{1+1}=2\]

Answer 10

so actual job is to compute
\[\frac{\frac{x+3}{x+1}-2}{x-1}\]

Answer 11

now there is no avoiding the algebra, we have to actually do the subtraction in the numerator, unless you want to multiply top and bottom by \(x+1\) to clear the fraction

lets go ahead and subtract, working only in the numerator

Answer 12

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

Answer 13

now we get to
\[\frac{\frac{-x+1}{x+1}}{x-1}=\frac{-x+1}{(x+1)(x-1)}\]
and the usual miracle occurs, one factor in the denominator will cancel since
\[\frac{-x+1}{x-1}=-1\] your final answer is
\[\frac{-1}{x-1}\]

Answer 14

my most sincere thankyou's i belive that is correct

Answer 15

or if you prefer
\[-\frac{1}{x+1}\]

Answer 16

all steps are there, if any are confusing let me know

Answer 17

yw