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Mathematics 51 Online
zarkam21:

Help on second part

zarkam21:

1 attachment
zarkam21:

Have a draw a graph and answer the questions that are being asked

mhchen:

Hmmm well for a) 1/x^2 it looks like this'|dw:1539925289394:dw| So the thing is, it's NOT continuous at x=0. f'(c) = 0 basically means that there's a slope of 0, or a straight line, somewhere. But here there isn't a straight line.

mhchen:

Do you understand what f'(c) = 0 means?

zarkam21:

yes but it also is asking if its continius on the [a,b] iinterval which is a=-1 and b =1

mhchen:

Yeah, and it's not continuous at x=0, which is between a=-1 and b=1. Do you see why?

zarkam21:

yes because there is not an ongoing function at that point

mhchen:

yeah, and it goes to infinity so we call that an 'infinite discontinuity' So that answers (a) right? we'll move on to (b)

zarkam21:

well it is also asking if it is differentiable

mhchen:

oh well...shoot

zarkam21:

A. If it is continous at [a,b] B. If it is differentiable on (a,b) C. Draw the graph D. Show if there either does or does not exist a point c in (a,b) such that f'(c)=0

zarkam21:

This is what needs to be answered for each question

mhchen:

Vocaloid how do you show if a function is continuous and differentiable?

Vocaloid:

uh like, visually, if a function is continuous it shouldn't have any jump/point discontinuities to be differentiable at a certain point the RH and LH limits need to be the same

zarkam21:

so its not continuous and its not differentiable

mhchen:

Yeah I guess you'd have to draw the graph, and write like "based on the graph, it's not continuous or differentiable at x=0 since it goes to infinity there"

zarkam21:

D. Show if there either does or does not exist a point c in (a,b) such that f'(c)=0

zarkam21:

theres not point right

zarkam21:

no*

mhchen:

Yeah there's not. There's no straight lines anywhere.

zarkam21:

okay next one

mhchen:

Next one is easy. 2x(x-1) a=0 b = 1 that's just a quadratic

zarkam21:

graph would be a parabola for this one

zarkam21:

right

mhchen:

And this time, f(a) = f(b) since f(0) = 0 and f(1) = 0 It's continuous and differentiable everywhere. And if you draw the graph |dw:1539926306171:dw| it looks like this

mhchen:

|dw:1539926345760:dw| see the straight line here?

zarkam21:

yes

zarkam21:

is that the tangent line

zarkam21:

just need to show if there is a point now

mhchen:

yeah. well it looks like it's at x=1/2 You can try to show that f'(1/2) = 0

mhchen:

First find the derivative of 2x(x-1) and then plug in 1/2. If it comes out to be 0, it works

zarkam21:

it does work !

zarkam21:

so make a point at (1/2,0)

mhchen:

Yep.

mhchen:

Do the same thing for c). it's a quadratic again so it should be differentiable and continuous then find out if there's any straight lines somewhere again. If you think there are, do the f'(c) thing again

zarkam21:

|dw:1539927036684:dw|

zarkam21:

@Vocaloid this would be the graph right

Vocaloid:

for c)? the vertex should be on (1,0) not where you have it

zarkam21:

|dw:1539927110184:dw|

Vocaloid:

good, then you'd look for a place where the derivative is 0 between x = 0 and x = 2

zarkam21:

1?

Vocaloid:

good

zarkam21:

so thde point would be (1,0)

zarkam21:

where it would show that there is apoint present

Vocaloid:

the question is asking you to prove that there's a point where f'(c) = 0 since you found a horizontal tangent line at x = 1 you've proved that f'(c) = 0 exists on the interval (a,b)

zarkam21:

OKaay i got that

Vocaloid:

if you want to prove it algebraically (I don't think you need to but whatever) you can calculate the derivative of f(x) at x = 1

zarkam21:

NO i don't need to

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