if f(x)=(x+3)^2 then

f(x+h)-f(x)/h = Ax+Bh+C

Find constants A, B, and C.

Question

if f(x)=(x+3)^2 then

f(x+h)-f(x)/h = Ax+Bh+C

Find constants A, B, and C.

anonymous · Answer

when i plug the first thing into the second everything cancels out

sasogeek · Answer

do you know about differentiation by partial fractions?

anonymous · Answer

dont think so. i just know how to do rate of change and how to take the derivative of something...

myininaya · Answer

what is $$f(x+h)$$ given that $$f(x)=(x+3)^2$$

anonymous · Answer

((x+3)^2) + h) ???

myininaya · Answer

you just replace x with x+h

sasogeek · Answer

what that means is that, (x+h) imply x, so where ever u see x, replace it with (x+h) :)

amistre64 · Answer

i think they should not use x in both cases; when x = a+h might be less confusing

sasogeek · Answer

true

anonymous · Answer

i think thats what i did and everything canceled out on the left side

amistre64 · Answer

if f(x)=(x+3)^2 then when f(a)  and f(a+h)

myininaya · Answer

amistre you want a function of x not a since both sides is in terms of x hehe :)
you can use your fancy a if you want to though

amistre64 · Answer

x = a ;)

amistre64 · Answer

we use substitution all the time in calculus and diffies

sasogeek · Answer

$$\large \frac{(x+h+3)^2-(x+3)^2}{h} $$

anonymous · Answer

hmm i see

myininaya · Answer

now plug multiply that top junk out and some stuff will zero out

sasogeek · Answer

$$\large \frac{(x+h+3)^2-(x+3)^2}{h}=Ax+Bh+C$$

myininaya · Answer

then you will be able to factor out in h on top

myininaya · Answer

then you will be like oh h/h=1
i don't have that crazy h on bottom anymore

anonymous · Answer

so i would get 1 on the left?

anonymous · Answer

if so i did it differently and still got 1

myininaya · Answer

$$\frac{(x+3+h)^2-(x+3)^2}{h}=\frac{(x+3)^2+2(x+3)h+h^2-(x+3)^2}{h}$$

myininaya · Answer

$$=\frac{(x+3)^2-(x+3)^2+2h(x+3)+h^2}{h}$$

myininaya · Answer

$$=\frac{0+2h(x+3)+h^2}{h}=\frac{2h(x+3)+h^2}{h}=\frac{h(2(x+3)+h)}{h}$$
do you see what i mean about the h/h thing now?

anonymous · Answer

yeah...

sasogeek · Answer

you'll have so many of the h's cancelling out, and the rest will be easy to figure out, equate your final equations to the right hand side, I hated partial fractions so i'll probably have trouble with it though

anonymous · Answer

could you tell me how you foiled this (x+h+3)^2 or whatever?

sasogeek · Answer

what do you mean by foil... the writing or how that expression came about?

anonymous · Answer

or simplify...

anonymous · Answer

if there was only two things in the paranthesis, you woul foil, but what would you do with this

asnaseer · Answer

you could take advantage of the fact that you have the difference of two squares in the expression to simplify first.

myininaya · Answer

$$(a+b)^2=a^2+2ab+b^2$$
$$(x+3+h)^2=(x+3)^2+2(x+3)h+h^2$$

sasogeek · Answer

$$\huge (a+b)^2=a^2+2ab+b^2 $$$$\huge a=x+3 $$$$\huge b=h $$

asnaseer · Answer

$$\frac{(x+3+h)^2-(x+3)^2}{h}=\frac{(x+3+h+x+3)(x+3+h-x-3)}{h}$$$$=\frac{(2x+6+h)h}{h}=2x+6+h$$

asnaseer · Answer

this uses:$$a^2-b^2=(a+b)(a-b)$$

sasogeek · Answer

Naseer I believe she was referring to " (x+h+3)^2 " which she stated lol :P but yeah what u said is true though, i didn't even notice it myself :P good to know :D

asnaseer · Answer

yes saso - I was just highlighting a simpler way of getting to the same answer that both you and myininaya got to

myininaya · Answer

whatever simpler 
my answer rules!

anonymous · Answer

me = he

sasogeek · Answer

always taking the shortcuts huh :P

asnaseer · Answer

:)

sasogeek · Answer

my bad bro, sorry lol

anonymous · Answer

and im still a little lost but if stare at this a little longer ill get it down in my head lol

sasogeek · Answer

sure :)

asnaseer · Answer

I can redo the steps from the beginning for you if it helps mario?

asnaseer · Answer

I know it can be confusing when you get lots of ideas thrown at you

anonymous · Answer

if it'll make me understand everything better.

anonymous · Answer

we just started doing all of this in class so its new material to me

sasogeek · Answer

i love this guys enthusiasm to study lol, keep it up mario :D

asnaseer · Answer

ok, we started with:$$f(x)=(x+3)^2$$using this we get:$$f(x+h)=(x+h+3)^2$$

asnaseer · Answer

therefore:$$f(x+h)-f(x)=(x+h+3)^2-(x+3)^2$$$$=(x+h+3+x+3)(x+h+3-x-3)$$$$=(2x+h+6)h$$

asnaseer · Answer

all ok so far?

anonymous · Answer

=(x+h+3+x+3)(x+h+3−x−3)
=(2x+h+6)h

dont understand this stuff

anonymous · Answer

what did you do to get from point a to point b

asnaseer · Answer

there I used the fact that:$$a^2-b^2=(a+b)(a-b)$$

asnaseer · Answer

where a in this case is x+h+3
and b is x+3

asnaseer · Answer

are you familiar with this rule?

anonymous · Answer

a little, yes

anonymous · Answer

how did you know which one is a and which one is b

asnaseer · Answer

$$a^2-b^2$$is compared to:$$(x+h+3)^2-(x+3)^2$$

asnaseer · Answer

so $a^2$ is the term on the left and $b^2$ is the term on the right

anonymous · Answer

ok i see i see

asnaseer · Answer

ok, so we got to:$$f(x+h)-f(x)==(2x+h+6)h$$divide both sides by h to get:$$\frac{f(x+h)-f(x)}{h}=2x+h+6$$

asnaseer · Answer

you are told that this also equals:$$Ax+Bh+C$$therefore:$$2x+h+6=Ax+Bh+C$$now just compare the coefficients of x and h to determine A and B and then C is just the constant term.

anonymous · Answer

that was a really good explanation. thanks a lot!

asnaseer · Answer

yw

asnaseer · Answer

and sorry to saso and myininaya for stepping on your shoes :(

anonymous · Answer

haha

myininaya · Answer

i still like my way better :)

asnaseer · Answer

I will let the lady take the high ground :)

anonymous · Answer

sasogeek, myininaya, and amistre, thank you as well :)