calc

[paradise!]

need help with this question.

I'm told f and g are differentiable and at x=5
f(5)=7
f ' (5)=-3
If f(7)=-1 and f'(7)=4 what can i conclude from the intermediate value theorem.?

Well i have an idea.

Since it is differentiable, it is continous. so all points f(5) to f(7) exist. therefore. all points f '(5) to f '(7) exist. But what can be told about their values? Am i correct?

And yes, i have looked at google. Many times.

[Jedi DC]

all y between 7 and -1 have an x value, not so much to this question.

[jord]

go on yahoo answers
someone on there will always be able to answer any question you have
it's helped me with homework a lot

[Nockm]

The only other thing I can grasp is that although between f(5) -> f(7) it goes from 7 -> -1, it doesn't just go down a single decreasing hill, it increases first and does some wavy shit then increases its way back to -1. I doubt that's the answer but that's my 0.0338999714 Turkish liras

[Stompa]

yo paradise, lemme hook u up with some of my calculus homework from high school, fucking 5 years ago.

http://i36.tinypic.com/wk0opz.jpg

http://i35.tinypic.com/343sqp0.jpg

maybe your answer's up in that?

[Fluffz]

But what can be told about their values?

At least one of them is zero?

E x : f(x)=0 | x e (5,7)
E x : f'(x)=0 | x e (5,7)
E x : sup f(x) > 7 | x e (5,7)
E x : max f(x) >= 7
E x : min f(x) < -1
E x : inf f(x) < -1 | x e (5,7)

Oh man i think this is finaly right now asdfg. Btw, It isnt.

[Zeebu]

id go with something along the lones of what fluffz said:

there is a least one number 'a' inside 5<a<7 where f(a)=0
there is a least one number 'b' inside 5<b<7 where f'(b)=0

the first one because it has to cross the y=0 line at least once

the second because the direction of slope changes at least once inside that range (finding your mins and maxes in this range)

[Nockm]

yo paradise, lemme hook u up with some of my calculus homework from high school, fucking 5 years ago.

http://i36.tinypic.com/wk0opz.jpg

http://i35.tinypic.com/343sqp0.jpg

maybe your answer's up in that?

Find x

[HandofDeath]

If this is the first year of Calc, I probably remember enough to answer.. What Calc class is it?

[DankNuggets]

i haven't had calculus in several years, but from a quick wikipedia refresher I'm guessing that it's basically saying what Zeebu said.

If it's continous, then for any "y" value (he used 0 for an example) within the endpoint values (7 to -1) there is at least one "x" value that it corresponds to. Also, it applies to the dervitive if it's continuously differentiable too, so there's a "dy/dx" between your initial and final y' values (and per your example 0 is between -3 and 4) You could also tell that it's a local minimum, don't know if that's part of it.

It might help you clear up things to wiki

"intermediate value therom" - i think the first and last sections are probably the closest to what you want

also "mean value therom" seems oftly familiar to me with questions like that.


EDIT: it also means that for any given y value between 7 and -1 there may be two x values for the same y value. don't know how to explain this well without a graph, but think of it like a parabola - there's two ways to get a particular y value (though it may be only for a small set of x values here)

[Vykromond]

need help with this question.

I'm told f and g are differentiable and at x=5
f(5)=7
f ' (5)=-3
If f(7)=-1 and f'(7)=4 what can i conclude from the intermediate value theorem.?

Well i have an idea.

Since it is differentiable, it is continous. so all points f(5) to f(7) exist. therefore. all points f '(5) to f '(7) exist. But what can be told about their values? Am i correct?

And yes, i have looked at google. Many times.

f takes all values between 7 and -1 (plus possibly values below -1) between x=5 and x=7
f has a local minimum between x=5 and x=7

pretty elementary stuff, pay more attention in class

[Fluffz]

why do we assume this function is continous again? what speaks against a graph of -1/x^2? because that would mean there is a x value without y value.

[Blocks]

why do we assume this function is continous again? what speaks against a graph of -1/x^2? because that would mean there is a x value without y value.

Since the function is differentiable, it's continuous as well.

[Fluffz]

Since the function is differentiable, it's continuous as well.

Says who? 1/x is is differentiable but not continuous.

[Vykromond]

you are terrible. 1/x is continuous. all differentiable functions are continuous, it's not even a hard proof

can you confine your shittiness to the gay bashing thread please