# Delta?

Show 40 post(s) from this thread on one page
Page 1 of 2 12 Last
• Aug 22nd 2008, 04:40 PM
CalcGeek31
Delta?
what does the symbol delta(the triangle symbol)x mean?
• Aug 22nd 2008, 04:59 PM
galactus
It is the 4th Greek letter and usually represents 'change in' x
• Aug 22nd 2008, 05:19 PM
CalcGeek31
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 PM
Spec
The problem is most likely to find the derivative of f in the point x=2.
• Aug 22nd 2008, 05:35 PM
Chop Suey
Quote:

Originally Posted by CalcGeek31
so what would I do if the problem is:

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

the equation:x^2 - 2x + 1

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:

$\lim_{\Delta x \to 0} \frac{f(2+\Delta x ) \color{red}- f(2)}{\Delta x}$
• Aug 22nd 2008, 05:35 PM
CalcGeek31
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 PM
Chop Suey
Let $h = \Delta x$

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

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)

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

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

Now simply replace and simplify. You'll find that you can cancel the h in the denominator after all.
• Aug 22nd 2008, 05:42 PM
CalcGeek31
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 PM
CalcGeek31
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 PM
Chop Suey
Quote:

Originally Posted by CalcGeek31
wait... I get down to lim(h-> 0) (2h + h^2 - 2) / h
then how do I cancel the denominator?

You made a mistake while simplifying:
$\lim_{h \to 0} \frac{f(2+h) - f(2)}{h}$

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

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

...
• Aug 22nd 2008, 06:00 PM
CalcGeek31
ahhh... k so my final answer is 2?
• Aug 22nd 2008, 06:02 PM
Chop Suey
Quote:

Originally Posted by CalcGeek31
ahhh... k so my final answer is 2?

EDIT: Yes, I made trivial mistake. You're correct.
• Aug 22nd 2008, 06:05 PM
CalcGeek31
$
f(2+h) = (2+h)^2 - 2(2+h) - 1
$

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

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

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

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

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

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

$\lim_{h \to 0} (h+2) = 2$
• Aug 22nd 2008, 06:11 PM
CalcGeek31
$
\lim_{h \to 0} \frac{4 + 4h + h^2 - 4 +2h - 1 +1}{h}
$
[
your +2h in here should be -2h...
Show 40 post(s) from this thread on one page
Page 1 of 2 12 Last