Help on second part
Have a draw a graph and answer the questions that are being asked
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.
Do you understand what f'(c) = 0 means?
yes but it also is asking if its continius on the [a,b] iinterval which is a=-1 and b =1
Yeah, and it's not continuous at x=0, which is between a=-1 and b=1. Do you see why?
yes because there is not an ongoing function at that point
yeah, and it goes to infinity so we call that an 'infinite discontinuity' So that answers (a) right? we'll move on to (b)
well it is also asking if it is differentiable
oh well...shoot
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
This is what needs to be answered for each question
Vocaloid how do you show if a function is continuous and differentiable?
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
so its not continuous and its not differentiable
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"
D. Show if there either does or does not exist a point c in (a,b) such that f'(c)=0
theres not point right
no*
Yeah there's not. There's no straight lines anywhere.
okay next one
Next one is easy. 2x(x-1) a=0 b = 1 that's just a quadratic
graph would be a parabola for this one
right
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
|dw:1539926345760:dw| see the straight line here?
yes
is that the tangent line
just need to show if there is a point now
yeah. well it looks like it's at x=1/2 You can try to show that f'(1/2) = 0
First find the derivative of 2x(x-1) and then plug in 1/2. If it comes out to be 0, it works
it does work !
so make a point at (1/2,0)
Yep.
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
|dw:1539927036684:dw|
@Vocaloid this would be the graph right
for c)? the vertex should be on (1,0) not where you have it
|dw:1539927110184:dw|
good, then you'd look for a place where the derivative is 0 between x = 0 and x = 2
1?
good
so thde point would be (1,0)
where it would show that there is apoint present
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)
OKaay i got that
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
NO i don't need to
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