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camoJAX Group TitleBest ResponseYou've already chosen the best response.0
http://www.purplemath.com/modules/fcnnot2.htm
 one year ago

Goten77 Group TitleBest ResponseYou've already chosen the best response.1
this means whereever there is a x in the original equasion that you substituate in x+h instead
 one year ago

Goten77 Group TitleBest ResponseYou've already chosen the best response.1
dw:1362544377090:dw
 one year ago

laurenn_marie7 Group TitleBest ResponseYou've already chosen the best response.0
@Goten77 so I got to 2x^2+4xh+2h^2+5x+5h+2 ... is that as far as I can get?
 one year ago

Goten77 Group TitleBest ResponseYou've already chosen the best response.1
were the directions to expand it?
 one year ago

laurenn_marie7 Group TitleBest ResponseYou've already chosen the best response.0
No just to find the value
 one year ago

Goten77 Group TitleBest ResponseYou've already chosen the best response.1
dw:1362545462751:dw
 one year ago

Goten77 Group TitleBest ResponseYou've already chosen the best response.1
oh then pluggign in x+h is all ya had to do
 one year ago

laurenn_marie7 Group TitleBest ResponseYou've already chosen the best response.0
oh.... so just 2(x+h)^2+5(x+h)+2 would be my answer?
 one year ago

laurenn_marie7 Group TitleBest ResponseYou've already chosen the best response.0
@Goten77
 one year ago

whpalmer4 Group TitleBest ResponseYou've already chosen the best response.0
That's the best you can do. You need to expand it, though. Usually this particular construct is in the context of \[\lim_{h\rightarrow0}\frac{f(x+h)f(x)}{h}\]and the subtraction of f(x) removes many of the terms and the h in the denominator cancels one or more, hopefully leaving you with a limit you can evaluate without too much pain and suffering. In the case of this problem, you would have \[\lim_{h\rightarrow0}\frac{2x^2+4xh+2h^2+5x+5h+2(2x^2+5x+2)}{h} = \]\[\lim_{h\rightarrow0}\frac{4xh+2h^2+5h}{h} =\lim_{h\rightarrow0}4x+2h+5 = 4x+5 \]
 one year ago

whpalmer4 Group TitleBest ResponseYou've already chosen the best response.0
And as you'll find out soon enough, \[\frac{d}{dx}[2x^2+5x+2] = 4x+5\]
 one year ago

laurenn_marie7 Group TitleBest ResponseYou've already chosen the best response.0
So confusing...so is it 4x+5 ? @whpalmer4
 one year ago

whpalmer4 Group TitleBest ResponseYou've already chosen the best response.0
Sorry, that was a preview of coming attractions :) No your answer is \[2x^2+4xh+2h^2+5x+5h+2\]just like you asked about.
 one year ago

laurenn_marie7 Group TitleBest ResponseYou've already chosen the best response.0
Gotcha! Thank you so much! What about for f(x)? @whpalmer4
 one year ago

whpalmer4 Group TitleBest ResponseYou've already chosen the best response.0
Unless there's a part of your problem statement that you didn't include or I didn't see, f(x) isn't part of your problem, only my digression about computing derivatives...
 one year ago

laurenn_marie7 Group TitleBest ResponseYou've already chosen the best response.0
No no f(x) of 2x^2+5x+2 @whpalmer4
 one year ago

whpalmer4 Group TitleBest ResponseYou've already chosen the best response.0
if \(f(x) = 2x^2+5x+2\) then \(f(x) = (2x^2+5x+2) = 2x^25x2\), no?
 one year ago

whpalmer4 Group TitleBest ResponseYou've already chosen the best response.0
Not to be confused with \(f(x) = 2(x)^2+5(x)+2 = 2x^25x+2\)
 one year ago
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