I'll just go over Instantaneous and average speed since no one seems to really have done that yet.

A wolf runs at y=93+2.3(t)^2 miles in t hours. What is it’s average speed during the first 5 seconds?

1. State the slope formula: (y2-y1)/(x2-x1)
2. Put the time you need to find into parenthesis or brackets: [0, 5] These are your two x’s.
3. Plug x1 into your formula: y=93+2.3(0)^2 = 93 93 is your y1
4. Plug x2 into your formula: y=93+2.3(5)^2=150.5 150.5 is your y2
5. Plug back into your slope formula: (150.5-93)/(5-0)
6. Simplify: 11.5 miles per hour
   
   We can just use the same one for instantaneous speed.

A wolf runs at y=93+2.3(t)^2 miles per hour.

1. What is the instantaneous speed for t=3?
   State the Instantaneous Speed formula:
   Lim f(t-h)-f(t)
   H->0 h
2. In the original problem, replace y with f(t): f(t)=93+2.3(t)2
3. Plug in: Lim 93+2.3(3-h)^2-93+2.3(3)^2
   H->0 h
4. Foil the parenthesis:
   Lim 93+2.3(9-6h+h^2)-93+2.3(9)
   H->0 h
5. Simplify:
   Lim (93+20.7 –13.8h+2.3h^2) – (93 + 20.7)
   H->0 h
6. Once you get it in it's simplest form, plug in for h.
   Lim 2.3(0)+27.6
   H->0
7. ANSWER: 27.6 miles per hour.

For what i don't understand..i thought i understood limits, but idk what my deal is this week.
When it says is the limit of f(x) or f(2) larger.. what exactly am i doing.. i think
its one of the first questions on the ch. 1 study guide.