The definition of continuity states a function is continuous if three conditions are met: f(c)
exists, f (x) = f (x) = f (c). The first condition is met, f(−2) exists, from part (b). The
second condition is violated; the left and right hand limits as x approaches −2 are not equal, so
the limit does not exist and the function is not continuous: g (x) = 5 and g (x)
= 5.
(d) Find the value of 3 x(9 − x^2 ) dx
Solve using u-substitution, in which u = 9 − x^2 and du = 2x dx. Thus, 3 x(9 − x^2 ) dx = 145.8.
- A particle moves with velocity v(t) = 9t^2 + 18t − 7 for t ≥ 0 from an initial position of s(0) = 3.
(a) Write an equation for the position of the particle.
The position function of the particle can be determined by integrating the velocity with respect
to time, thus s(t) = ∫v(t) dt. For this problem, s(t) = ∫ (9t^2 + 18t − 7) dt = 3t^3 + 9t^2 − 7t + C.
Since we are given the initial position, s(0) = 3, plug that in to solve for C. Thus, C = 3 and the
equation for the position of the particle is s(t) = 3t^3 + 9t^2 − 7t + 3.
(b) When is the particle changing direction?
The particle changes direction when the velocity is zero, but the acceleration is not. In order to
determine when those times are, set the velocity equal to zero and solve for t: v(t) = 9t^2 + 18t −
7 = 0 when t = and t = − . Since, the time range in question is t ≥0, we can ignore t = −.
Then, take the derivative of the velocity function to find the acceleration function, as (v(t)) =
a(t). For the given v(t), a(t) = 18t + 18. Check that the acceleration at time t is not zero by
plugging into the acceleration function: a(t) = 30. Therefore, the particle is changing direction
at t = because v(t) = 0 and a(t) ≠ 0
(c) What is the total distance covered from t = 2 to t = 5?
The distance covered is found by using the position function found in part (a). Determine the
position at t = 2 and subtract it from the position at t = 5. From part (b), we know that the
object does not change direction over this time interval, so we do not need to find the time
