Asked by John
                I have 3 calc problems that i have not been able to do please help me... i've been on here yesterday as well but no one was able to assist me with these particular ones...
1. If f(x)=5x^3–2/x-4
find f'(x)____
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? ____
3. f(x)=2x^2+7x+3/ sqroot of x
Find f'(x)
i got ((x^(-1/2))(4x+7)+1/2x^(-3/2)(2x^2+7x+3))/x
            
            
        1. If f(x)=5x^3–2/x-4
find f'(x)____
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? ____
3. f(x)=2x^2+7x+3/ sqroot of x
Find f'(x)
i got ((x^(-1/2))(4x+7)+1/2x^(-3/2)(2x^2+7x+3))/x
Answers
                    Answered by
            Damon
            
    1. If f(x)=5x^3–2/x-4
find f'(x)____
=============================
I assume you mean
(5 x^3 -2)
----------
(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
==============================
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
----------------------------------------
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]
--------------------------------------
x
(sqrt x/x) (6 x^2 + 7 x -3)
6 x^(3/2) + 7 x^(1/2) - 3 x^-(1/2)
======================================
i got ((x^(-1/2))(4x+7)+1/2x^(-3/2)(2x^2+7x+3))/x
                    Answered by
            John
            
    I don't exactly know why but all of the answers were wrong. 
    
                    Answered by
            Damon
            
    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
    
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
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