Question

HELP ASAPPP

Answer 1

@Kevin

Answer 2

For the first question

\[lim_{x \rightarrow 2^{-}} f(x)\] implies the value of the function when x approach 2 from left hand side

Answer 3

For the second question
The expression for velocity is \[\frac{ds(t)}{dt}=\frac{d}{dt}(-6-2t)\]


Find the expression of velocity from here and find instantaneous velocity at t=2 by replacing t by 2 in velocity expression

Answer 4

What is the difference between 2^- and 2 ?

Answer 5

@inkyvoyd

Answer 6

@phi

Answer 7

are you asking how to take the derivative of
-2 - 6 t
?

Answer 8

no, I am asking the two attachments

Answer 9

ok. The first is asking for the limit as x->2-
which means if you start to the left of 2 (i.e on the negative side of 2)
what does f(x) approach?

do you see the line for f(x) ?
what "y" value does it have "near" 2 (but on the left side of 2)

Answer 10

do you see the f(x) line? what y value does it have at x=1 (for example)?

Answer 11

do you see the f(x) line? what y value does it have at x=1 (for example)?

can you answer that , or do you not understand how to read the graph ?

Answer 12

ok, a bit less than 2, I think.

any way, x-> 2- means as x (on the left side of 2 i..e. for numbers like 1.5 or 1.9)
the y value comes down to some number a bit bigger than 1)
that is the limit.
only one of the choices seems it could be the right number.

Answer 13

For the second problem, they want you to find the derivative of
-2 -6t

did you learn how to take a derivative of a "constant" -2 ?
how about the derivative of -6 t ?

Answer 14

@phi ?

Answer 15

for the first one

Answer 16

yes, 1.3 is the "left limit"

Answer 17

@phi
can you explain me further about 2^- ?
I still don't understand

Answer 18

\( 2^-\) is a way to say "approach 2 from the left side of 2 i.e. 1.9, 1.99,...)
as opposed to starting above (bigger than) 2
the limit also means we *never* reach x=2 (so we never evaluate f(2) )

Answer 19

hmm... ok, how if I change it become limit x -> 2

Answer 20

@phi

Answer 21

For the second problem, they want you to find the derivative of
-2 -6t

did you learn how to take a derivative of a "constant" -2 ?
how about the derivative of -6 t ?


*Kevin x->2 means approach 2 from both sides. The limit does not exist if we don't get the same number from either side. As in this problem, the limit for x-> does not exsit as
lim x-> 2- is 1.3 and lim x->2+ is 4

Answer 22

So if it has a negative power, we should read from the left to right
If it has a positive power, we should read from the right to left?

Answer 23

Are you saying you have not studied how to take a derivative ?
or that you studied, but you are confused about how to do it?

Answer 24

*Keven yes, but it's not really a "power". It's just a short-hand way to say what you said.

Answer 25

I am just confused on this specific one..

Answer 26

do you know how to find the derivative of
f(x) = -2 - 6x
?

Answer 27

@phi
Thx!

Answer 28

First, velocity is change is distance divided by change in time
In your problem, where they give you position
the change in position/change in time will be the velocity
in this case, it will be the slope of the line
y= -6x -2 (if we used y for p, and x for t)

for background , see
https://www.khanacademy.org/math/calculus-home/taking-derivatives-calc/derivative-as-a-limit-calc/v/calculus-derivatives-1-new-hd-version

Answer 29

A point of interest the limit as x-->2+ is 4 (approaching from the righr side)
There are different limits for the 2 different approaches. Due to this we say the limit as x---> 2 ( with no sign after it) does not exist

Answer 30

@welshfella so that would be my answer??

Answer 31

No The question was the limit as x --->2- The answer to that is 1.3
I only added that comment for your info ( you might be asked limit as x --> some value in another similar question)

Answer 32

@welshfella @triciaal

Answer 33

as ph1 explained velocity = ds/dt
so u need to find the derivative of -2-6t

Answer 34

the derivative of a constant is zero so the -2 disappears
to find the derivative of -6t to use the generak form

if y = ax^n , dy/dx = anx^(n-1)

Answer 35

so for -6t (which is same as - 6t^1)
a = -6 and n = 1 ( and x = t of course)
plug thos into the general formula

Answer 36

listen to phi

Answer 37

????

Answer 38

one combined answer would not help you to understand the concept.
all of them are using same method
if you learned it differently in school then go with that one.
it depends on you.

Answer 39

@welshfella please help

Answer 40

@welshfella hello?? you're still here

Answer 41

@welshfella

Answer 42

@mathmate @jhonyy9

Answer 43

@iwanttogotostanford do you still need help? Looking back at these answers, it looks like you've got it figured out? :)

Answer 44

@sweetburger @johnweldon1993 will be able to help further if you need it. :-)

Answer 45

@hanamalik no I still need help I don't know the answer directly

Answer 46

@jackthegreatest @Loser66 please help!!

Answer 47

I need a new post. there is no time to go over a long post like this.