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Calculus I                                                  Quiz #6                            March 23, 2004

Name____________________             R.  Hammack                          Score ______
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(1) Find the following derivatives.

(a)      f(x) = 3x^3 + 1/x + tan(x) - 4

    f(x) = 3x^3 + x^(-1) + tan(x) - 4

    f ' (x) = 9x^2 + (-1) x^(-1 - 1) + tan(x) - 0 = 9x^2 - 1/x^2 + sec^2(x)


(b)     d/dx[ x sec(x) ] = (1) sec(x) + x sec(x) tan(x) = sec(x) + x sec(x) tan(x)



(c)     d/dx[(x^2 sin(x))/(x + 1)] = (d/dx[x^2 sin(x)] (x + 1) - x^2 sin(x) d/dx[x + 1])/(x + 1)^2 =((2x sin(x) + x^2cos(x)) (x + 1) - x^2 sin(x))/(x + 1)^2




(2)  An object moves on a straight line in such a way that its distance from a stationary point at time  t seconds is given by the function f(t) = t^3 - 6t^2 + 6t + 2 feet.  At what time(s)  t  does the object have a velocity of -3feet per second?

Velocity at time t is f ' (t) = 3t^2 - 12t + 6 feet per second.
We seek the time at which velocity is -3 feet per second.
Such a time t must satisfy the following equation.
f ' (t) = 3t^2 - 12t + 6 = -3
Solving this, we get
3t^2 - 12t + 6 = -3
3t^2 - 12t + 9 = 0
t^2 - 4t + 3 = 0
(t - 1) (t - 3) = 0

Thus at times t =1 and t = 3 the object has a velocity of -3 feet per second.