Question

For the function, find and simplify
[f(x + h) − f(x)]/h. (Assume h ≠ 0.)



how do I plug in f(x)=9x^2
into
[f(x + h) − f(x)]/h

Answer 1

So we're trying to set up our difference quotient:
\[\Large\rm \frac{\color{orangered}{f(x+h)}-\color{royalblue}{f(x)}}{h}\]

They provided us with this function:\[\Large\rm \color{royalblue}{f(x)=9x^2}\]Understand the color coding?
see where we plug our function in?

Answer 2

yes

Answer 3

\[\Large\rm \frac{\color{orangered}{f(x+h)}-\color{royalblue}{9x^2}}{h}\]Ok good, that takes care of that part then.
Now we need to deal with the \(\Large\rm f(x+h)\)

Answer 4

ok.

Answer 5

I been confused since I was not given f(x+h)

Answer 6

So we're taking our function f,
and instead of plugging x into it, we're plugging x+h in.\[\Large\rm f(\color{green}{x})=9(\color{green}{x})^2\]In the same way that we might plug 2 in for x,\[\Large\rm f(\color{green}{2})=9(\color{green}{2})^2\]We can also plug in x+h for our x.\[\Large\rm f(\color{green}{x+h})=9(\color{green}{x+h})^2\]Understand how that works? :d

Answer 7

Why is it that for f(x+h) we plug (x+h) into 9x^2
but for f(x) we left it as 9x^2.

Answer 8

I am confused on how f(x+h) is 9(x+h)^2

Answer 9

Function notation can be a little tricky :)

here is our function:\[\Large\rm f(~~)=9(~~)^2\]Just think of it like this for a moment.

Answer 10

ok

Answer 11

So when we say that this is a function of x,
we're "plugging" x into our function,\[\Large\rm f(\color{orangered}{x})=9(\color{orangered}{x})^2\]When we plug something different in,\[\Large\rm f(\color{orangered}{x+h})\]We get something different out,\[\Large\rm =9(\color{orangered}{x+h})^2\]

Answer 12

When it was written as a function of x,
the x's are x. :P

When we instead plug in x+h,
we x's are replaced with x+h's.

Still a lil confusing? :d

Answer 13

oh wow ok I like that. Makes sense

Answer 14

So here is our function `evaluated at` (value plugged in) x+h,
\[\Large\rm \color{orangered}{f(x+h)=9(x+h)^2}\]

Answer 15

so here is what i get when I actually plug f(x) and f(x+h) into the equation without simplifying:

[9(x+h)^2 - 9x^2] / h

Answer 16

Ok great!

Answer 17

Gotta be careful with your simplification.
Remember how to expand a binomial?

Answer 18

and when I have simplified I came up with this:

[9(x^2+2xh+h^2)-9x^2] / h
then :
(9x^2+18xh+9h^2-9x^2) / h

Answer 19

then :
(18xh+9h^2) / h

Answer 20

can I then factor out 9h and be left with :
[9h(2xh+h)] h
?

Answer 21

oh I cant factor out 9h can I?

Answer 22

Yes you can.
But there is one further simplification step from there.

Do the numerator and denominator share any factors?
Can we do anything about that?

Answer 23

yes, H

Answer 24

ok : so.....
if I did that before pulling out 9h I would have
18x+9h
?

Answer 25

Yessss, very good \c:/

Answer 26

You can pull out the 9 from there if you want, but it's not necessary.

Answer 27

ok I have one mroe quick question and I will take it from here if I am correct ok

Answer 28

from what you have taught me I will attempt to plug in the next two functions on my own

Answer 29

I have f(x) = 7x2 − 9x + 3
so [f(x + h) − f(x)] / h

so

I would have

{[7(x+h)^2-9(x+h)+3]-[7x^2-9x+3]} / h

Answer 30

?

Answer 31

f(x) 7x2 − 9x + 3

Answer 32

Yessss, very good on your setup.

Remember that in order to drop these brackets [7x^2-9x+3],
you'll have to distribute the negative to each term.

Answer 33

yes I did and I simplified and I ended up with this (7x^2+14xh+9h+9x+7h^2) / h

Answer 34

then if i simplified it with H being a common denominator I came up with this:
7x^2+14x+9+9x+7h

7x^2+23x+7h+9

, is this correct?

Answer 35

Before you do our division, taking the h's out,
You shouldn't be left with any non-h terms.
Just something to keep in mind.

So I think we made a little mistake somewhere.
Let's see where.

Answer 36

So from here,
\[\Large\rm \frac{7(x+h)^2-9(x+h)+3-7x^2+9x-3}{h}\]Expanding gives us:\[\Large\rm \frac{7x^2+14xh+h^2-9x-9h+3-7x^2+9x-3}{h}\]Combining like-terms:\[\Large\rm \frac{\cancel{7x^2}+14xh+h^2\cancel{-9x}-9h+\cancel{3}\cancel{-7x^2}+\cancel{9x}\cancel{-3}}{h}\]

Answer 37

Early on you had a -9(x+h) and I think you accidentally turned it into 9x+9h.

It should have become -9x-9h

Answer 38

oh wow, I should have ruled out the common denominator beforehand like you are showing, that would save a lot of work.
and yes I caught that thank you:)

Answer 39

well after erasing opposing "factors" ez: -3 , +3

Answer 40

ok... I see what you did

Answer 41

I can simplify h from the numerator, and it ends up canceling out with the H from the denominator
?right?

Answer 42

leaving me with 14xh +h -9

Answer 43

Yes, everyone in the top loses an h.
Woops I think you forgot to take an h out of 14xh

Answer 44

yes you're right, this is why its bad to work and do my homework at the same time. hah

Answer 45

haha XD

Answer 46

at least I have the main issue down , how to plug f(x) and f(h+x) into an equation, now I just have to go slowly through the steps and always go back and recheck those steps. :)

Answer 47

So I plugged in our final answer, and was told it was wrong.

Answer 48

good good good :)
your basic algebra steps seem very good.
I'm glad we got past that function stuff without too much trouble.

Answer 49

Wrong? +_+
Lies...
Sec I'll rework it on paper :3

Answer 50

me too
even though you will prbably have it done way before I do

Answer 51

This is what the function looks like?\[\Large\rm f(x)= 7x^2 − 9x + 3\]

Answer 52

\[\Large\rm f(x)= 7x^2 -9x + 3\]

Answer 53

yes

Answer 54

and f(x+h) - f(x) all over h

Answer 55

14x-9+7h

That's what I'm coming up with ^
Ahhh I think I didn't look at your answer closely enough last time >.<

Answer 56

Oh oh oh that was my fault.
I didn't distribute the 7 to the h^2.

that term in the numerator should had been a 7h^2.

Answer 57

I eneded up with this: 14xh+7h^2-9h

Answer 58

Before taking an h out? Good.

Answer 59

hahaha i didnt see your psot, but I missed it too so its ok, its a long equation and its late a night,. its forgivable by me lol

Answer 60

yes

Answer 61

then : 14x+7h-9

Answer 62

and I don't believe I can simplify that anymore, so This will be my final answer

Answer 63

yes!!!! now it is correct!! thank you@zepdrix for all your help! :)

Answer 64

Yay team \c:/