Posted by John on Sunday, April 27, 2008 at 8:30pm.
1. If f(x)=5x^3–2/x-4
find f'(x)____
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I assume you mean
(5 x^3 -2)
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(x-4)
(x-4)(15 x^2) - (5 x^3-2)(1)
----------------------------
x^2 - 8 x + 16
15 x^3 -60 x^2 - 5 x^3 +2
-------------------------
x^2 - 8 x + 16
10 x^3 -60 x^2 +2
-----------------
x^2 - 8 x + 16
what i got was (4x^2+7x-12)/x^8
2. The position function of a particle is given by s=t^3–4t^2–7t t>=0
where s is measured in meters and t in seconds. When does the particle reach a velocity of 10 m/s? ____
===========================
s' = 3 t^2 - 8 t - 7 = 7
3 t^2 - 8 t -14 = 0
t = [ 8 +/- sqrt (64 +84) ] /6
= [8 +/- 12.17]/6
=3.36
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3. f(x)=2x^2+7x+3/ sqroot of x
Find f'(x)
sqrt (x) (4x+7) - (2x^2 + 7 x +3)(1/2x)sqrt x
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x
2 x sqrt (x)(4x+7) - (2 x^2+7x+3) sqrt x
---------------------------------------
x
(sqrt x)[8 x^2 +14 x - 2 x^2 -7 x - 3]
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x
(sqrt x/x) (6 x^2 + 7 x -3)
6 x^(3/2) + 7 x^(1/2) - 3 x^-(1/2)
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i got ((x^(-1/2))(4x+7)+1/2x^(-3/2)(2x^2+7x+3))/x
I don't exactly know why but all of the answers were wrong.
2. The position function of a particle is given by s=t^3–4t^2–7t t>=0
where s is measured in meters and t in seconds. When does the particle reach a velocity of 10 m/s? ____
===========================
s' = 3 t^2 - 8 t - 7 = 10
3 t^2 - 8 t -17 = 0
t = [ 8 +/- sqrt (64 + 204) ] /6
= [8 +/- 16.4]/6
=4.94