Question

thanks

Answer 1

This deals with distance, all right? So let's let that distance (or 'closeness') be represented by \(\large \Delta x\) ok?

Answer 2

i follow

Answer 3

Now, our x here in this case is c = 2.
We want the difference of \[\Large g(c+\Delta x)<0.001\]right?

Answer 4

Wait, sorry, something missing
\[\Large \left|g(c+\Delta x) \color{red}{-8}\right|<0.001\]

Answer 5

ok

Answer 6

Let's take the positive case first:
\[\Large g(\color{blue}2+\Delta x) - 8 < 0.001\]
can you solve for \(\large \Delta x\)?

Answer 7

g+x < 6.001?

Answer 8

... Let me help you out a little...
since \(2+\Delta x \ne 2\)
Then
\[\Large g(2+\Delta x) = 3(2+\Delta x) + 2\]
carry on from here, please...

Answer 9

im slightly confused on how you translated that, but if you distribute the 3(2 +x) then you get 6+ 3x+ 2

Answer 10

Your function?
\[\Large g(x) = \begin{cases}3x+2\qquad x\ne 2\\\\3\qquad \qquad \ \ \ x=2\end{cases}\]

Answer 11

so then that gives 8+3x?

Answer 12

not x, per se, \(\Delta x\) but I understand that you can't type that delta normally, right?
Let's use h instead.
\[\Large g(2+h)=\color{red}? \]

Answer 13

2g+ g(h)? This is where im utterly confused from thelecture notes

Answer 14

You don't understand how function notation works? D:
g is not a factor to distribute, it is simply the name of the function.
For example, if we have
\[\Large f(\color{red}x) = \color{red}x^2 - 4\color{red}x+4\]
then\[\Large f(\color{red}2)= \color{red}2^2 - 4(\color{red}2) + 4 = \color{blue} 0\]
\[\Large f(\color{red}{x+h}) \ne f(x) + f(h) \qquad \qquad \color{red}{!!!!!}\]\[\Large f(\color{red}{x+h}) = (\color{red}{x+h})^2 - 4(\color{red}{x+h}) + 4\]
You understand?
Just replace all the x's in g(x) with x+h.

Go.

Answer 15

And by 'go' I mean tell me what g(2+h) is equal to.

Answer 16

I understand regular notation. I dont understand how the above (2+h) or (2+ delta x) is simplified down if its already had input? 2+h would be equal to the bottom line of the function no?

Answer 17

I just want you to work out (and simplify) g(2+h)
for now.
come on... g(x) = 3x + 2
g(2+h) = ?

Answer 18

we're solving g(2+h) for the x delta correct, so simply 2...?

Answer 19

No.
You were so close earlier. (close is just a nice way of saying 'wrong' ^_^)
Nonetheless, I already explained it... just replace all the x's in g(x) with 2+h and then simplify.

Answer 20

so 3(2+h)+2 is what we did earlier with the 8+3 delta x, no?

Answer 21

8+3? no.
Again... (sorry if I'm pushing you, trust me, this is one of the simpler maths, so you really have to master this ^_^)

Answer 22

can we backtrack then

Answer 23

Sure.
Replace the x's in g(x) = 3x + 2
with 2+h
and then simplify
There is an unknown h, so do NOT expect the simplification to be just a simple number.

Answer 24

ok so, 3(2+H)+2, simplify that, 3*2=6, 3*H=3h, so 6+3h+2, 8+3h,

Answer 25

Okay, good.
Now, this is what we want:
\[\Large g(2+h) -8 <0.001\]
We already know g(2+h) = 8+3h.
So...\[\Large 8+3h - 8 < 0.001\]care to solve for h? ^_^

Answer 26

so add 8 to .001, then subtract 8, so 3h < .001, divide by 3, h < 3.33E^-4(somehow I think this is wrong again)

Answer 27

I'd just stick to \[\Large \frac{0.001}{3}= \frac1{3000}\]

Answer 28

ok

Answer 29

so .0003

Answer 30

Well, approximately, yes.

Answer 31

That means when c is less than 0.00033333... units far from 2, the value of g(x) will differ from 8 by no more than 0.001.

Job...well, almost well done, so promise me you'll practice with your algebra ^_^

Answer 32

alright, much appreciated, ima have to reread this but thanks

Answer 33

Note that it's necessary to assume h is positive here.

Answer 34

but that's just a technicality

Answer 35

alrighty will do