Posted by Becky on Monday, August 13, 2012 at 6:12pm.
Steve already went through this with you in
http://www.jiskha.com/display.cgi?id=1344880951
Look at
F(v) : (600^3/2 + 20250)/v
what we are saying is that we suspect that that term would be 600v^(3/2)
I got (-150v^(3/2) + 40500)/v^3
As I told you in an earlier post, I would have simplified the original to
f(x) = 600 v^(1/2) + 20250/v and I would not have used the quotient rule, but our answer should come out the same .
My way: ....
f(x) = 600 v^(1/2) + 20250 v^-1
f'(x) = (1/2)600v^(-1/2) - 20250v^-2
= 300 v^(-1/2) - 20250v^-2
which is what you would get if we simplified your answer.
then f''(x) = -150v^(-3/2) + 40500v^-3
Your answer should have been
(-150v^(3/2) + 40500)/v^3
Ok but the power 5/2 is wrong? Should have been power 3/2?
I keep on getting the answer 5/2.. I don't know which part of my working is wrong
I do not want the simplified answer by the way...I need the answer as an expression only
based on my question, i retyped...
F(v) : (600v^3/2 + 20250)/v
F'(v) : (300v^3/2 - 20250)/v^2
F"(v) : (-150v^5/2 + 40500)/v^3
Reiny, isn't your answer suppose to be
f(x) = 600 v^(3/2) + 20250 v^-1,
instead of f(x) = 600 v^(1/2) + 20250 v^-1 ??
1/2 power and 3/2 power there's a difference....
hi i think i know what you mean already.
my answer for F"(v)
(-150v^5/2 + 40500v)/v^4
i simplified further, it became
(-150v(v^3/2 - 270))/v^4
cancel off the common factor v
it became (-150(v^3/2 - 270))/v^3
Am i correct??
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