Question

For the function f(x)=4/x+3. Find the average rate of change of f from 1 to x.

Answer 1

really from 1 to \(x\) ?

Answer 2

i guess it is
\[\frac{1}{x-1}\int_1^x\frac{4}{t+3}dt\]

Answer 3

anti derivative is \(4\ln(t+3)\) so integral would be
\[\frac{1}{x-1}4\left(\ln(x+3)-\ln(4)\right)\]

Answer 4

So how would I find the rate of change?

Answer 5

what class is this for? it it calculus?

Answer 6

College Algebra

Answer 7

then my answer is not the one you are looking for

Answer 8

are you sure it is the rate of change from 1 to \(x\) and not from 1 to some number

Answer 9

Blahhh so hard for me unfortunately

Answer 10

I am sure

Answer 11

okay then this is what you need to compute
we can do it, it is not that hard
\[\frac{f(x)-f(1)}{x-1}\]

Answer 12

this is the change in \(f\) over the change in \(x\) just like the slope of a line
now comes some annoying algebra

Answer 13

They gave me options for the answer but I dont know how to put fractions on this site

Answer 14

no problem, we can work out the answer
first off we need \(f(1)=\frac{4}{1+3}=\frac{4}{4}=1\)

Answer 15

so now we have to compute
\[\frac{f(x)-1}{x-1}\]

Answer 16

we get
\[\frac{\frac{4}{x+3}-1}{x-1}\] as a first step

Answer 17

a) -1/x+3
b) 1/x+3
c) 4/x(x+3)
d) 4/(x-1)(x+3)

Answer 18

now multiply top and bottom by \(x+3\) to clear the fraction
you get \[\frac{\frac{4}{x+3}-1}{x-1}\times \frac{x+3}{x+3}\]
\[=\frac{4-(x+3)}{(x-1)(x+3)}\]
\[=\frac{4-x-3}{(x-1)(x+3)}\]
\[=\frac{1-x}{(x+3)}\]

Answer 19

typo on the last line

Answer 20

it should be
\[=\frac{1-x}{(x-1)(x+3)}\]

Answer 21

then since \(\frac{1-x}{x-1}=-1\) the
final answer
is
\[\frac{-1}{x+3}\]

Answer 22

Thank you!!! ;)