What does the fact that h(0)= 0 have to do with being a critical point?I changed the critical point to 0, though, because the integral with the bounds [0,0] would equal 0.

The derivative of h(x) is f(x). The critical points of h are where f= 0, not where h= 0. Note that for x< 5, f(x)< 0 so h(x) is decreasing while for x> 5, f(x)> 0 so h(x) is increasing. There is a relative minimum of h(x) at x= 5. The opposite happens at x= 11. f(x) goes from positive to negative, so h(x) goes from increasing to decreasing. There is a relative maximum of h(x) at x= 11.

h(12) is NOT -1 and h(0) is NOT -4. Those are the values of f(x), not h(x)!

It's not that difficult to find h(x) explicitely. For x= 0 to 3, f(x)= -4 so h(x)= -4x. For x= 3 to 7, f(x)= 2x- 10 so [tex]h(x)= x^2- 10x+ 21 (the constant, 21, is so that ). For x= 7 to 12, f(x)= -x+ 11 so (the constant, -52.5, is so that ). Calculate the values at the critical points, x= 5 and x= 11, and at the endpoints x=0 and x= 12 to see what the absolute maximum and minimum are.

That doesn't even make sense. The problem asked for a1. Find the time when the particle has an acceleration of 9 meters per squared second.

-I solved this and got s(6)=147time, not a value of s. Do you mean that your answer is 6 seconds?

Okay, what2. Find a function to represent the position of the particle at any time t.

-s(t)=s(0) + integral of u^2-3u-4 du [0,t]isthat integral?

That says that when t= 0, v= -4 which has nothing to do with "when its velocity is zero." You want to solve v(t)= 0 for t, then put that value of t into the position function. Since the velocity function is quadratic, there will probably be3. Find the position of the particle when its velocity is zero.

-I was unsure so I just solved v(0)=-4.twosolutions.

Surely you know what "positive" and "negative"4. Determine when the velocity of the particle is positive and when it is negative.

-This is the problem I had no clue on.mean! Solve the inequalities v(t)> 0 and v(t)< 0. Hint: v(t) can change from positive to negative only where v(t)= 0- look at (3) again.