BEC'S methods questions

[bec]

ohhhhhhh i get it! i was differentiating the original equation and subbing in x=9
clearly i completely missed the point of what you were doing hahaha, but i completely understand now
thanks

[bec]

is there any way of including domain restrictions in graphs on the ti-89?
eg. y = 2x  where x>3
     y = 7    where x<3

[unknown id]

Function domains may be restricted on the TI 83/84:

Eg. Plot →
•   The operations < , ≤ , > , ≥ , 'and' , 'or' are obtained through the TEST, LOGIC menu.
•   Enter Y1=(x^2)(x>0).

Eg. Plot →
•   Enter Y1=(2x^2 − 5)(x≥−1 and x≤3)

Function domains may also be restricted on the TI 89:

Eg. Plot
•   The operations < , ≤ , > , ≥ , 'and' , 'or' are obtained through the MATH, TEST menu.
•   Enter Y1=2x|x>3
•   Remember the '|' symbol after the function, which has its own button on the keypad.

[bec]

thanks unknown id!

Let f: R-->R where f(x) = 2x   
                                  x² + a  x>0


Show that there is a value for a for which f is continuous and differentiable at every point of R.

I couldn't work it out, and according to the solutions, a= 1. How does that work though, isn't that a discontinuation at x=0?

[droodles]

no, it keeps going through 0, so you sub in a point i think

[Mao]

is there any way of including domain restrictions in graphs on the ti-89?
eg. y = 2x  where x>3
     y = 7    where x<3

to apply a constraint, simply add "| x>3" or "| x>3 and x<5"
or if you are using the newest version of OS (3.10), you can simply input "|3<x<5"
to check your OS, go to your apps screen and "about", if you are Titanium, it is inside Menu (F1)

[Mao]

thanks unknown id!

Let f: R-->R where f(x) = 2x   
                                  x² + a  x>0


Show that there is a value for a for which f is continuous and differentiable at every point of R.

I couldn't work it out, and according to the solutions, a= 1. How does that work though, isn't that a discontinuation at x=0?

you are right, , as that is neither continuous nor differentiable.

in fact, without introducing a in the quadratics, that question cannot be done.

the only possibility is:

[bec]

yeah that's what i thought
i hate wrong answers in text books!

so just to clarify, the same hybrid graph i described before (but with y = x² + ax is differentiable at x=0 only when a=2 because...

When f(x) = 2x, f'(0) = 2
When f(x) = x² +ax, f'(0) = 2(0) +a

Both equations need the same value for f'(0)


or have i just invented a theory?

[Mao]

yes bec, that is exactly right

the formalised definition:

Continuity
for a function , it is continuous at iff:
 
exists

exists



or in plain english: the left and the right must join up at the same point.


Differentiability:
for a function , it is differentiable at iff:

is continuous at



or in plain english: the left and the right must join up smoothly

[Glockmeister]

Here's another way of explaining differentiablility

[IMG]http://img407.imageshack.us/img407/6075/apex1gn0.th.jpg[/img]

[IMG]http://img184.imageshack.us/img184/2472/apex2uh7.th.jpg[/img]

[bec]

hahaha

[bec]

f(x) = x(x-3)²
a) Find the co-ords of the points at which f'(x)=0   (1,4) and (3,0)
b) Find the equation of the straight line joining the points at which f'(x)=0  (y=-2x+6)
c) Hence, find the point of intersection of f and the stright line that lines between the two points of zero gradient, whose equation was found in part c. (2,2) What do you notice? [/u]


i got the actual answers right...but i dunno, what should i notice?

[ed_saifa]

its the midpoint?

[Matt The Rat]

Midpoints of the two co-ords of f'(x)=0

[bec]

good point

so what do i conclude from that?