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Posted by on Sunday, April 27, 2008 at 8:30pm.

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

  • Calculus please help - , Sunday, April 27, 2008 at 9:10pm

    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

  • Calculus please help - , Sunday, April 27, 2008 at 9:18pm

    I don't exactly know why but all of the answers were wrong.

  • misread this one - , Monday, April 28, 2008 at 1:45pm

    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

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