I have AP Calculus AB summer work, and I'm having trouble working out some of the problems. I'll probably be posting a few others as well. Any help would be appreciated!

x 0 1 2 3 4 5 6
f(x) -10 -8 -1 0 5 7 2

Using the table of values given above, find the average of change for f on the interval (2,5) and write an equation for the secant line passing through the corresponding points.

Average of change for f on the interval (2,5) = df/dx = [7-(-1)]/(5-2) = 8/3

That's then the gradient of your line, m=8/3.

Then just solve for y=mx+c (derive c) based on the two points, y=7 when x=5, and y=-1 when x=2.

gives c= y-mx = 7-8/3*5 = ... (based on the first point)

Or, c = y-mx = -1-8/3*2 = ... (based on the second point)

Thanks for the quick reply, but can you explain what you did there?

Thanks!

Yep, you're basically just calculating the gradient between the two points where x=2 (f=-1) and x=5 (f=7), so you can ignore all the other points/data.

I've also added some hints on the next part, above.

So, would the line be y=(8/3)x+(19/3)?

Just came across another one.

Given that h(x) = (tan^-1)[1-(pi)(x)] is a composite function of the form h(x)=f(g(x)), find f and g.
[I think may be referring to the previous problem where f(x)=x/(x-1), and g(x)=1/(x-1), but I have no idea.]

Just put x=2 and then x=5 back into your equation to make sure the line runs through those two points. Sorry for not giving you a direct answer, but you should do some of the thinking for yourself.