Question

help please ...
http://prnt.sc/b2o8wi

Answer 1

@ganeshie8

Answer 2

Start by finding the derivative

Answer 3

@zepdrix

Answer 4

uh ohhh :O -5

Answer 5

yesh D:

Answer 6

idk what to do tbh but doesnt the mean theorem state
1. has to be polynomial ? and how do you know if its continous ?

Answer 7

1. has to be continuous
2. has to be differentiable? I think? maybe we don't need this.

Answer 8

Notice that if your interval had included the value t=3,
then we would have a big problem, right?

Answer 9

yea bc itll make it undefined ?

Answer 10

Yes.
Since t=3 wasn't included in the interval,
we don't have to worry about any discontinuity issues.
Yes, there is a discontinuity at t=3, but it's not in the interval [-2,0].

So the theorem should work okkie dokkie :)

Answer 11

So our theorem tells us that there is at least one tangent slope (in the interval) equivalent to the secant slope which connects the end points of the interval.

Answer 12

ohh okay so im still confuse though .. is there an algebraic way to find out if its continuous or not ?

Answer 13

Mmmm no, I don't think so.
You just have to remember what type of "math things" create "bad stuff" for us.
Examples would be:

Square root can be bad because of negative numbers.
Dividing by stuff can be bad because of dividing by 0 at some particular place.
Ummm what else.. I dunno, just gotta watch for stuff like that.

Oh log might be another one, 0 and negative numbers bad.

Answer 14

I should be careful the way I say "bad stuff" though,
because not all of those are discontinuity issues,
in fact only the dividing by 0 is.

The others are domain issues.
The negative values are not even in the domain of the square root functions.

Maybe that's a little technical though.

Answer 15

So to solve this problem, know what to do?

Derivative will give us equation for `tangent slope`.
Slope formula on the original equation will give us `value of the secant slope`.
And then set those equal to one another.

Answer 16

That moment when you thought you ran out of chicken nuggets,
but there was one hiding in the corner.
Scoreeee! :O

Answer 17

ahh okay but im not sure if im wrong if its contious though but isnt usually when its 1/x or 1/x ^2

Answer 18

because i graph this fucntion for 7 and i got something like this

Answer 19

Maybe this is what's confusing you.
We don't need TOTAL CONTINUITY.
We just continuity on our interval.

So if your discontinuity is located `outside of the given interval`,
then you just completely ignore it.

Answer 20

ohhh

Answer 21

like, sure the function 1/x has some problems at x=0.
But if we were given the interval [1,5] then we can confidently say we have continuity on that interval because 0 isn't inside it.