[mattf]

ok, seriously...

5a) for g(0) you just plug in 0 for x, so you have the integral from -3 to 0 of f(t) which is just the area under the curve on that interval (which is a trapezoid).

for g'(0), you take the derivative of g(x) first which is f(x)(1) - f(-3)(0), or just simply f(x). plug in 0 (look at the graph).

5b) g(x) is the integral of f(t), so f(t) is the derivative of g(x). any time your derivative changes sign from + to -, you have a maximum. look at the graph at x=3.

5c) you start the same way you did with part b, only looking for changes in sign from - to + (x=4). however, since it's checking for an absolute (rather than relative min), you have to check the functional value at whatever point you get as well as at your endpoints. i remember this part took a while.

5d) when the second derivative (f'(t) in this case) has a change in sign (equals 0), you have a point of inflection (x=1). keep in mind that a curve is not differentiable at a cusp (x=-3, for instance).

6a) they give you a differential equation, and remember that a derivative is the slope of a tangent line at a given point. plug in the coordinates for the given points to get the slopes for each point and sketch. for example, at (1,2), the slope would be 1, so you draw a 45 degreeline through (1,2).

6b) this will be easy to do if you can see the slope field.

6c) use separation of variables to integrate and solve for y. then plug in 0 for x and 3 for y to solve for C (constant of integration) and substitute back in.

hope that clears things up for you.