Last minute DESPERATE q's OMFG!!!

[2007vce]

HOW DO U DO THESE QUESTIONSS??? ITS HARDDD


ALSO!!!!!!!!!!!
when do u use R=MA  -->  (x)i+(y)j = ma (the vector crap)
and when do u just resolve, resolve horizontally, resolve vertically
? I DUNNO, WHEN U USE WHICH ONE? ANY???

THANKSSSSSSSS

[Collin Li]

A point of inflection is when a double derivative crosses the x-axis. In other words, when double derivative is zero and dy/dx is not zero (so that there is no turning point, and this ensures the double derivative crosses the x-axis).

This is painful to the eye, and I'm on the phone. I'll get back to this later.

[brendan]

f'' = second derivative
f' = first derivative

where f'' is positive it is said that the curve is concave up
where f'' is negative it is said that the curve is concave down

An inflection point is a point on the curve where the curvature (concavity) changes.

Therefore, at the inflection point, f'' changes sign

If f'' changes sign (from either +ve to -ve or vice versa), and if f'' is continuous, then by the intermediate value theorem, you know that f'' crosses the x axis. Let x = a be where f'' crosses the x-axis. Thus f''(a) = 0. Therefore, f' does not change

[Collin Li]

f'' = second derivative
f' = first derivative

where f'' is positive it is said that the curve is concave up
where f'' is negative it is said that the curve is concave down

An inflection point is a point on the curve where the curvature (concavity) changes.

Therefore, at the inflection point, f'' changes sign

If f'' changes sign (from either +ve to -ve or vice versa), and if f is continuous, then by the intermediate value theorem, you know that f'' crosses the x axis. Let x = a be where f'' crosses the x-axis. Thus f''(a) = 0. Therefore, f' changes sign at x = a

Don't confuse her with your university first semester Maths knowledge! My explanation does not go into the gory details, but it's all you need to know for Specialist (and any further probably is too hard to understand). Not good when your exam is tomorrow, haha.

[Collin Li]

The more general case for a point of inflection is when the double derivative changes sign. Brendan highlighted (on MSN) the counter-example to what I said originally with: y = x^(1/3)

[Ahmad]

If f'' changes sign (from either +ve to -ve or vice versa), and if f is continuous, then by the intermediate value theorem, you know that f'' crosses the x axis. Let x = a be where f'' crosses the x-axis. Thus f''(a) = 0. Therefore, f' does not change

I find this a bit weird. You need f'' to be continuous not f to apply the intermediate value theorem, by virtue of the fact that f'' exists f is continuous.

[brendan]

f''[0] = 0, but clearly f'

• changes sign.

if f'' is zero, then that implies that at that point f' is not changing.

[Ahmad]

f''[0] = 0, but clearly f'

• changes sign.

if f'' is zero, then that implies that at that point f' is not changing.

Yeah but you're checking for sign changes on the sides of f'.

[Ahmad]

POI discussion has come up a few times on here already. I've already written about it lol.  :x

[Collin Li]

Heh, it's a maths-debate.  :lol: (Had to say it)

[brendan]

f''[0] = 0, but clearly f'

• changes sign.

if f'' is zero, then that implies that at that point f' is not changing.

Yeah but you're checking for sign changes on the sides of f'.

but you don't need f' to change sign for a point of inflection.

[Ahmad]

Heh, it's a maths-debate.  :lol: (Had to say it)

Healthy discussion, not debate!

[Collin Li]

Heh, it's a maths-debate.  :lol: (Had to say it)

Healthy discussion, not debate!

Wanking is healthy.

[Ahmad]

I thought you were trying to show that dy/dx does not change sign.

Anyway here is how I'd do it:
f''[a] = 0 and f'' changes sign at a. Therefore f'[a] is either a minimum or maximum and therefore does not change sign.

[brendan]

I thought you were trying to show that dy/dx does not change sign.

Anyway here is how I'd do it:
f''[a] = 0 and f'' changes sign at a. Therefore f'[a] is either a minimum or maximum and therefore does not change sign.

what about the function f(x) = x^(1/3)