31 replies to this topic

[Melchoire]

I had this question my Calculus final.

Sketch a graph that fits these conditions:
f(1)=0
f(-1)=0
f'(x) > 0 for all x

I really struggled with this one. Doesn't make any sense to me at all.

[jcrdude]

This may not be possible, but it's a sketch that fulfills the given parameters.

[phalkon]

i think jcr is correct.

its been a while, but that's roughly right.

[unworthy]

I loved my calculus class back when I did it. My teacher was awesome and would sit there with us for 45 minutes doing a question. So time consuming I do not miss it. Did your final only have like 10-12 questions? We did not have enough time to do more than that many

[vendetta.inc]

Wouldn't f(1) and f(-1) equaling zero mean that their y values are equal to zero? Not vertical asymptotes. A vertical asymptote would be present at these values if you have limits towards 1 and -1, where the limit approached infinity.

The function very well could still have vertical asymptotes. But not at 1 and -1.

The third parameter relates that the slope of f(x) is always greater than zero throughout the function. Something like jcr drew fulfills this.

Edited by vendetta.inc, 18 December 2009 - 11:00 AM.

[Melchoire]

This may not be possible, but it's a sketch that fulfills the given parameters.

Nice try but when it says f'(x)>0 it implies that the function is differentiable for all values of x.

[jcrdude]

Nice try but when it says f'(x)>0 it implies that the function is differentiable for all values of x.

It is. When you differentiate, you get the following graph for f'(x)

[vendetta.inc]

It is. When you differentiate, you get the following graph for f'(x)

What is f'(0) then? It's an asymptote in your sketch, meaning there isn't a true value.

[Melchoire]

It is. When you differentiate, you get the following graph for f'(x)

Right and it wouldn't be differentiable at 0.

[Vexage]

This is giving me bad flashbacks to my days of calculus. Barely squeezed by.

[vendetta.inc]

Right and it wouldn't be differentiable at 0.

I'm trying to think of how this is even possible. Roots at 1 and -1, with positive slopes throughout. I'm lost.

[Melchoire]

I'm trying to think of how this is even possible. Roots at 1 and -1, with positive slopes throughout. I'm lost.

Yeh, but there's gotta be a way to do it. =P

[Drakonid]

Would a line which is = to the X axis work for that?

Edit: nevermind, just re read that, I was talking nonsense...

Edited by SilverRage, 19 December 2009 - 07:57 PM.

[PeonYourLawn]

i can confirm it was put on the test as a joke, you could probably state the rules of differentiability

[vendetta.inc]

Yeh, but there's gotta be a way to do it. =P

And thus why I hate calculus...

[MrUnknownGuy]

Oh yeahhhhhhhhhhhh this is possible.

[Cinder]

I think it's possible xD

Edited by Cinder, 20 December 2009 - 06:15 PM.

[artificial]

If you think about it logically than f'(x) > 0 means for all values of x, the curve of f(x) is increasing, or has a positive gradient. How then is it possible that it crosses the x-axis twice? The answer? It must have an asymptote, thus jcr is correct. If it didn't, it would have to be a parabola, which wouldn't satisfy those conditions.

[Melchoire]

If you think about it logically than f'(x) > 0 means for all values of x, the curve of f(x) is increasing, or has a positive gradient. How then is it possible that it crosses the x-axis twice? The answer? It must have an asymptote, thus jcr is correct. If it didn't, it would have to be a parabola, which wouldn't satisfy those conditions.

Perhaps but when I asked one of my TAs during the exam he said that "f'(x)>0" implies that it's differentiable for all x. I guess i'll have to wait and see on Jan. 4

[vendetta.inc]

If you think about it logically than f'(x) > 0 means for all values of x, the curve of f(x) is increasing, or has a positive gradient. How then is it possible that it crosses the x-axis twice? The answer? It must have an asymptote, thus jcr is correct. If it didn't, it would have to be a parabola, which wouldn't satisfy those conditions.

How is the curve increasing when there is an asymptote? The value at such an area is undefined, thus the curve is neither increasing, or decreasing.

Perhaps but when I asked one of my TAs during the exam he said that "f'(x)>0" implies that it's differentiable for all x. I guess i'll have to wait and see on Jan. 4

Your TA is correct.

[jcrdude]

It pleases me that this topic is still going. I look forward to seeing the "correct answer"

[Ender]

I'm pretty confident that one of the math geniuses on codex will pull through.

[Melchoire]

Lol, I'm still waiting for Joe to drop by. Or maybe he did and didn't want to admit he doesn't know it

[artificial]

Without an asymptote it should be impossible, lol.

If it has a root at (-1, 0), then if it keeps increasing throughout its domain it should never come down to the x-axis again.

[ShadowLink64]

I'm supposedly one of the "math" people here, and I can't figure this out for the life of me. The question must be written incorrectly.

I mean, the function cannot possibly have a positive gradient everywhere if the value of the function is the same for two different x values. There has to be a discontinuity in the derivative....

Without an asymptote it should be impossible, lol.

If it has a root at (-1, 0), then if it keeps increasing throughout its domain it should never come down to the x-axis again.

Exactly.