Question

How do you go to from the first form to the second y=4x/(x+3) at (-5,10)

[4(x+h)/(x+h)+3]-[4x/(x+3)] all over h
to this
12h/(x+h+3)(x+3)/h

Answer 1

lots of algebra

Answer 2

can you help?

Answer 3

@satellite73

Answer 4

help please

Answer 5

it is
\[f(x)=\frac{4x}{x+3}\] right?

Answer 6

i tried adding just H
I tried factoring out 1/h

Answer 7

yes at (-5,10)

Answer 8

oh you want it at a point, not in general
so you want
\[\frac{f(x)-10}{x-5}\]

Answer 9

make that
\[\frac{f(x)-10}{x+5}\]

Answer 10

want to find slope at the point and equation for tan line

Answer 11

f(x+h)-f(x)/h

Answer 12

ok so you either want this
\[\frac{f(x+h)-f(x)}{h}\] with no particular point, you can plug in the numbers later, or you want the specific one
\[\frac{f(x)-10}{x+5}\]

Answer 13

the second will be easier, the first is more general
you can plug in \(-5\) once you find it

Answer 14

I just want to know how they got 12h on top

Answer 15

ok

Answer 16

that =30?

Answer 17

lets do it in general
takes more algebra is all
first of all, lets ignore the \(h\) in the denominator, and once we compute
\[f(x+h)-f(x)\]we can divide by \(h\)

Answer 18

ok

Answer 19

\[f(x+h)=\frac{4(x+h)}{x+h+3}-\frac{4x}{x+3}\] is a start now we have to subtract

Answer 20

\[\frac{4(x+h)(x+3)-4x(x+h+3)}{(x+h+3)(x+3)}\] is step 1
leave the denominator in factored form

Answer 21

ok

Answer 22

all you did was multiply x+h+3 to nominator of second fraction?

Answer 23

no sorry cross multiply

Answer 24

i did this
\[\frac{a}{b}-\frac{c}{d}=\frac{ad-bc}{bd}\]

Answer 25

to have common denominator

Answer 26

right, that is the only way to add or subtract fractions

Answer 27

ok

Answer 28

now just plug -5 for x?

Answer 29

now we multiply all that mess out in the top\[\frac{4(x+h)(x+3)-4x(x+h+3)}{(x+h+3)(x+3)}\]
\[\frac{4x^2+4xh+12x+12h-4x^2-4xh-12x}{(x+h+3)(x+3)}\]

Answer 30

which is 12

Answer 31

actually it is \(12h\) but when you divide by \(h\) you are left only with the \(12\) up top

Answer 32

so slope at -5 is 3

Answer 33

12/(x+3)^2

Answer 34

that leaves\[\frac{12}{(x+3)^2}\]

Answer 35

right

Answer 36

at \(x=-5\) you get \(3\)
it would have been lots less algebra to do it the other way

Answer 37

and the line is

Answer 38

you got the slope \(m=3\) and the point \((-5,10)\) point slope formula gets it

Answer 39

just learning whats in book, not very good at math really cant even remember the easy stuff

Answer 40

y=3x+25

Answer 41

guess what? it is all algebra
lots of lots of algebra

Answer 42

but in a week you will be able to find the derivative of \(\frac{4x}{x+3}\) much quicker

Answer 43

@satellite73
comp dropped out thanks for help. I hope so.

Answer 44

yw

Answer 45

so that other way works the same?

Answer 46

if you do it the other way you just get a number, namely 3
it is easier, but doing it this way you get a formula that you can use for any number, not just \(-5\)
you will notice that we did not use \((-5,10)\) anywhere in the computation

Answer 47

which would be -10 -10/x-5

Answer 48

it would be
\[\lim_{x\to -5}\frac{f(x)-10}{x+5}\]

Answer 49

or
\[\lim_{x\to -5}\frac{\frac{4x}{x+3}-10}{x+5}\]

Answer 50

except for the tan line

Answer 51

which would be 0/0

Answer 52

right for that we used it, but once you have a general formula you can find the slope for any \(x\) so you can find the tangent line at any point

Answer 53

so you still have to do the algebra

Answer 54

yes, of course
once you get \(\frac{0}{0}\) it just means you have more work to do

Answer 55

so yo ucan use the other form for any point and sub y for f(x)

Answer 56

*can't

Answer 57

\[\lim_{x\to -5}\frac{\frac{4x}{x+3}-10}{x+5}\]
\[=\lim_{x\to -5}\frac{{4x-10(x+3)}}{(x+5)(x+3)}\]\[=\frac{-6x-30}{(x+3)(x+5)}\]
\[=\frac{-6(x+5)}{(x+3)(x+5)}=\frac{-6}{x+3}\] put in \(x=-5\) get \(3\)

Answer 58

nice