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 February 21st, 2017, 01:32 AM #1 Member   Joined: Feb 2017 From: East U.S. Posts: 40 Thanks: 0 Please teach me how to do this The limit represents f '(c) for a function f(x) and a number c. Find f(x) and c. lim (Δx-->0) [7-8(1+Δx)]-1(-1)/Δx So the story is... my teacher assigns this as a homework problem that's gonna turn up on our exam tomorrow, but without going over a single problem with a triangle thing in it. I looked it up and apparently it's a "change" symbol? Please, I really just need the steps written down on how to solve this so I can at least try to learn a little before my next exam in about 11 hours. I'm willing to try to learn, and not just looking for the answer, but I have no time... ANY advice would be great, thanks!
 February 21st, 2017, 02:49 AM #2 Senior Member   Joined: Feb 2016 From: Australia Posts: 1,519 Thanks: 506 Math Focus: Yet to find out. The 'little triangle' is an uppercase delta in Greek. And yes, it usually means 'change in'. When you see $\Delta x$, you can say 'change in x'. Although this by itself doesn't mean much. I can't really make out what your function is supposed to be due to some peculiar bracketing. $\displaystyle \lim\limits_{\Delta x \rightarrow 0} \dfrac{7 - 8(1 + \Delta x)}{\Delta x}$ or, $\displaystyle \lim\limits_{\Delta x \rightarrow 0} 7 - 8(1 + \Delta x) +\dfrac{1}{\Delta x}$??? Or something else...
February 21st, 2017, 03:46 AM   #3
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 Originally Posted by Joppy The 'little triangle' is an uppercase delta in Greek. And yes, it usually means 'change in'. When you see $\Delta x$, you can say 'change in x'. Although this by itself doesn't mean much. I can't really make out what your function is supposed to be due to some peculiar bracketing. $\displaystyle \lim\limits_{\Delta x \rightarrow 0} \dfrac{7 - 8(1 + \Delta x)}{\Delta x}$ or, $\displaystyle \lim\limits_{\Delta x \rightarrow 0} 7 - 8(1 + \Delta x) +\dfrac{1}{\Delta x}$??? Or something else...
Sorry, I don't know how to use the website... But

lim
Δx-->0

In the numerator of the function, I have "[7-8(1+Δx)]-(-1)" (there are brackets for some reason)

And in the denominator, I just have a "Δx"

Hopefully this clarified everything.

Last edited by nbg273; February 21st, 2017 at 04:35 AM.

 February 21st, 2017, 04:48 AM #4 Math Team   Joined: Dec 2013 From: Colombia Posts: 7,151 Thanks: 2390 Math Focus: Mainly analysis and algebra $$f'(c)= \lim_{\Delta x \to 0} \frac{f(c+\Delta x) - f(c)}{\Delta x}$$ $f(c+\Delta x) = 7-8(1+\Delta x)$ and $f(c) = -1$. By inspection, you can then suggest a value for $c$ and an expression for $f(x)$.
February 21st, 2017, 05:03 AM   #5
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 Originally Posted by v8archie $$f'(c)= \lim_{\Delta x \to 0} \frac{f(c+\Delta x) - f(c)}{\Delta x}$$ $f(c+\Delta x) = 7-8(1+\Delta x)$ and $f(c) = -1$. By inspection, you can then suggest a value for $c$ and an expression for $f(x)$.
Sorry, I'm still confused on what f(x) is... I'm bad at math...

 February 21st, 2017, 05:29 AM #6 Math Team   Joined: Jan 2015 From: Alabama Posts: 2,959 Thanks: 801 It looks like f(x) is intended to be f(x)= 7- 8x. Then $f(1)= 7- 8(1)= -1$ and $f(1+ \Delta x)= 7- 8(1+ \Delta x)= 7- 8- 8\Delta x$. So $f(1)- f(1+ \Delta x)= -1- (-1- 8\Delta x)= 8\Delta x$. That is, for all non-zero $\Delta x$, $\frac{f(1)- f(1+\Delta x)}{\Delta x}= \frac{8\Delta x}{\Delta x}= 8$. So what is the limit as $\Delta x$ goes to 0?
February 21st, 2017, 06:17 AM   #7
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 Originally Posted by Country Boy It looks like f(x) is intended to be f(x)= 7- 8x. Then $f(1)= 7- 8(1)= -1$ and $f(1+ \Delta x)= 7- 8(1+ \Delta x)= 7- 8- 8\Delta x$. So $f(1)- f(1+ \Delta x)= -1- (-1- 8\Delta x)= 8\Delta x$. That is, for all non-zero $\Delta x$, $\frac{f(1)- f(1+\Delta x)}{\Delta x}= \frac{8\Delta x}{\Delta x}= 8$. So what is the limit as $\Delta x$ goes to 0?
Ohhh, I see now. Thanks for your help!

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