Question

A particle moves along the x–axis so that at time t its position is given by x(t) = (t+1)(t–3)3. For what values of t is the velocity of the particle increasing?

Answer 1

I'll assume that's an exponent at the end there:

$x(t) = (t+1)(t-3)^3$

The first derivative gives the velocity.  The second derivative gives the acceleration; increasing velocity is the same as positive acceleration.  So we want to find when the second derivative is positive.

Let's see if we can use $d(uv) = u\, dv + v\, du$ to avoid multiplying this out.

$x'(t) = 3 (t+1)(t-3)^2 + (t-3)^3$

That worked; let's do it again:

$x''(t) = 3( 2(t+1)(t-3) + (t-3)^2) + 3(t - 3)^2$

$x''(t) = 3(t-3) ( 2(t+1) + (t-3) + (t - 3) ) = 3(t-3)(4t-4)=12(t-1)(t-3)$

That's a nice parabola.  It's zero or negative for $1 \le t \le 3$ so positive everywhere else:

Answer: Increasing velocity when t < 1 or t > 3