Calculus Limit and Function Help?

Have a final exam on Monday and for some reason I can't figure out these older problems. Please help and show me how you've come to the answer.

Thanks

If an arrow is shot upward on the moon with a velocity of 58 m/s, its height in meters t seconds later is given by h=58t-0.83t^2.

a) Find the average velocity over the given time intervals.

1. [1,2]

2. [1, 1.5]

3. [1,1.1]

4. [1,1.01]

5. [1,1.001]

b) Estimate the instantaneous velocity when t=1.

And the second problem.

Guess the value of the limit (if it exists) by evaluating the function at the given numbers (correct to six decimal places).

lim x^2-2x/x^2-x-2

x--> -1

x= 0,-0.5,-0.9,-0.95,-0.99,-0.999,-2,-1.5,-1.1,-1.01,-1.001

Thanks I really appreciate it. Points to whoever can help me!

3 Answers

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  • 1 decade ago
    Favorite Answer

    For part (a), evaluate the h function for each of the t-values in 1 through 5.

    t = 1, 1.001, 1.01, 1.1, 1.5, 2

    h = 57.17, 57.226339, 57.733317, 62.7957, 85.1325, 112.68

    The average velocity is the slope from point to point.

    i.e. the average velocity for t = [1, 2]

    v = (112.68 - 57.17)/(2 - 1) = 55.51

    Do that for each interval.

    The results will be:

    2. v = 55.925

    3. v = 56.257

    4. v = 56.3317

    5. v = 56.339

    For part (b), you just make a guess based on what you see the numbers above approaching. I would guess 56.34.

    In the second problem, just substitute the x-values into the function (x^2 - 2x)/(x^2 - x - 2).

    As the x-values get closer to -1, the functional values get closer to the value of the limit.

    When you determine the values on this one, you will see that it approaches negative infinity from the right and positive infinity from the left. Since the left and right limits do not agree, the limit does not exist.

  • Anonymous
    1 decade ago

    Well to start we know that the first derivative is the velocity.

    h(t) = 58t-0.83t^2

    so

    h'(t) = 58-1.66t

    (makes sense because the +58 corresponds to the initial velocity of throwing it up at 58 m/s)

    This derivative here gives you the instantaneous rate of change, so for part b you can just plug 1 in for t.

    Now as far as part a, its finding the secant line between the two points. Now forget my derivative talk up there for a second and we will use the Average Rate of Change formula.

    remeber Δ means change in, so it is the size of our interval.

    Δy/Δx = [f(x+Δx)-f(x)]/Δx

    so just plug in with your intervals of x (replace with t for you :) )

    = [f(1+1)-f(1)]/1 = [ (58*2-.83*2^2) - (58*1-.83*1^2) ] /1

    = 55.51

    So simply do that with all of your intervals, changing the Δx with the new smaller intervals.

    im tired hope this helps!

  • Anonymous
    4 years ago

    This seems nasty, yet its quite now no longer problematical. Plug in 0 into x. We get sin(2cos0)/5sec0. cos0=a million, so 2cos0=2. 5sec0=5/cos0. Cos0=a million, so we get 5. Dividing those 2 factors, we get sin2/5. To do those issues continuously first plug in x= even with sort it techniques. This time there grew to alter into now no longer something nasty in touch, in basic terms problem-unfastened trig.

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