what does the symbol delta(the triangle symbol)x mean?

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- Aug 22nd 2008, 04:40 PMCalcGeek31Delta?
what does the symbol delta(the triangle symbol)x mean?

- Aug 22nd 2008, 04:59 PMgalactus
It is the 4th Greek letter and usually represents 'change in' x

- Aug 22nd 2008, 05:19 PMCalcGeek31
so what would I do if the problem is:

(f(2+delta(x))+ f(2)) / delta(x)

the equation:x^2 - 2x + 1 - Aug 22nd 2008, 05:26 PMSpec
The problem is most likely to find the derivative of f in the point x=2.

- Aug 22nd 2008, 05:35 PMChop Suey
Delta, as Galactus said, represents change in x. In your case, it represents the change of x as it approaches zero. This is the derivative at the point x = 2. Also, you made a small typo:

$\displaystyle \lim_{\Delta x \to 0} \frac{f(2+\Delta x ) \color{red}- f(2)}{\Delta x}$ - Aug 22nd 2008, 05:35 PMCalcGeek31
Im not sure I understand...

my problem is f(x) is equal to the second equation... now evaluate the first equation... - Aug 22nd 2008, 05:40 PMChop Suey
Let $\displaystyle h = \Delta x$

$\displaystyle \lim_{h \to 0} \frac{f(2+h) - f(2)}{h}$

Our objective here is to get rid of the h in the denominator so we can find the limit. Now, let's evaluate f(2+h) and f(2)

$\displaystyle f(2+h) = (2+h)^2 - 2(2+h) - 1$

$\displaystyle f(2) = (2)^2 - 2(2) - 1$

Now simply replace and simplify. You'll find that you can cancel the h in the denominator after all. - Aug 22nd 2008, 05:42 PMCalcGeek31
ohhh... now that makes perfect sense... it takes the form of f(x+h) thank you

I also have a similar problem in which there is no change in x. does this mean I work it all the way up until I do the limit? - Aug 22nd 2008, 05:53 PMCalcGeek31
wait... I get down to lim(h-> 0) (2h + h^2 - 2) / h

then how do I cancel the denominator? - Aug 22nd 2008, 05:58 PMChop Suey
- Aug 22nd 2008, 06:00 PMCalcGeek31
ahhh... k so my final answer is 2?

- Aug 22nd 2008, 06:02 PMChop Suey
- Aug 22nd 2008, 06:05 PMCalcGeek31
$\displaystyle

f(2+h) = (2+h)^2 - 2(2+h) - 1

$

from that... -2(h) = -2 h + 4h - Aug 22nd 2008, 06:09 PMChop Suey
$\displaystyle \lim_{h \to 0} \frac{f(2+h) - f(2)}{h}$

$\displaystyle \lim_{h \to 0} \frac{(2+h)^2 - 2(2+h) - 1 - (-1)}{h}$

$\displaystyle \lim_{h \to 0} \frac{4 + 4h + h^2 - 4 -2h - 1 +1}{h}$

$\displaystyle \lim_{h \to 0} \frac{4h + h^2 -2h}{h}$

$\displaystyle \lim_{h \to 0} \frac{h^2 +2h}{h}$

$\displaystyle \lim_{h \to 0} \frac{h(h+2)}{h}$

$\displaystyle \lim_{h \to 0} (h+2) = 2$ - Aug 22nd 2008, 06:11 PMCalcGeek31
$\displaystyle

\lim_{h \to 0} \frac{4 + 4h + h^2 - 4 +2h - 1 +1}{h}

$[

your +2h in here should be -2h...