Solution:

The two keys here are (a) it repeats over and over (so we're pretty sure a trig function is involved) and (b) There are lots of sharp corners in the graph, (so we're pretty sure absolute value is involved, too.) What I'm going to try to do, then, is turn this graph back into one I recognize, then try to go forward:

• See all the sharp corners? That's what happened when absolute value took all of the parts of the graph that were under the x-axis and flipped them up. How would I undo that? By taking lots of parts and flipping them back; see what happens when you take all of the small humps and flipping them back under the x-axis (Go ahead; do it!) All of the humps go away, and what's left is ...
• something that looks like sine; however, it's amplitude is 3 instead of 1, and it goes from -2 up to 4. From your trig background, you can figure out that the equation of this graph is y = 1 + 3sin(x); so
• The equation of the graph above is y = |1+3sin(x)|.

• The average velocity found using s(t) between the times of t = 3 and t = 5 is about -1/2.
• The instantaneous velocity found using s(t) at t = 3 is about 2.
• The instantaneous velocity found using s(t) at t = 5 is about -3.
• The maximum height of the object is about 5 ft.

Hint: Try drawing the graph of such a thing first, then try to find the equation of something like it.

Solution:

Before I try to draw it, I'll try to reason a little:

• Since the secant slope between the two points is -1/2, and the 'run' between them is 2 units, the 'rise' between them must be -1. So, the point when t = 3 is 1 unit higher than when t = 5.
• Since the slope of the tangent line at t = 3 is positive and the slope of the tangent line at t = 5 is negative, the graph must be going up when t = 3 and down when t = 5.
• Since the graph is going up when t = 3 and down when t = 5, the maximum must happen between the times of t = 3 and t = 5.
• (By now you probably know that I'm looking for a quadratic function, which is the easiest one I can find that goes up and then comes down.)
• The only problem at the moment is where to put the maximum; it can't go halfway between 3 and 5 (else the slope at 5 would be the same size as the slope at 3, only negative.) Since the slope at 5 is bigger, I'm gonna figure it's further away from the top; so for the moment I'm gonna try having the top happen at about 3.75; one parabola which opens down that has the top at (3.75, 5) is y = -(x-3.75)²+5.
• Checking: The y-value at 3 is about 4.43, the y-value at 5 is about 3.43 (so that part is okay.) The slope of the tangent line at x = 3 is about 1.5, the slope of the tangent line at x = 5 is about -2.5, both of which are gonna have to be good enough for now! (If I wanted to fix that, I would have to make the graph steeper at both x = 3 and x = 5, which I can do by putting a constant next to the quadratic term, like y = - k(x-3.75)²+5.

Solution:

Okay, I know I'm putting the solution ahead of the problem, but I thought I should tell you how I'm gonna go about it; for each of the criteria listed in the original problem, I'm gonna give you how I think of that. (If you have a better way, please let me know and I'll gladly substitute your way for mine!)
Problem: Try drawing a single function y = f(x) which has the following properties simultaneously:

• The function values y = f(x) exists everywhere, except at x = -1, x = 1, and x = 4.

  Think: The function is gonna have a dot everywhere, except at the three listed points (it probably means something weird is going to happen at those places, too!)

• The limit

  exists everywhere, except at a = -1, and a = 4.

  Think: The only places where you're going to have to lift up your pencil and move to another height and stay there is when x = -1 and when x = 4.

• The function is continuous everywhere, except at x = -1, x = 1, x = 3 and x = 4.
  

  Think: The only places where you're going to have to pick up your pencil are at x = -1, x = 1, x = 3, and x = 4.

• f(x) is negative only when 0 < x < 1.5, and when x > 4
  

  Think: The graph is only under the x-axis when x is between 0 and 1.5, and to the right of when x = 4.

• The limit

  Think: Way to the left, the y-values on the graph are getting very close to the line y = 0 (so that way to the left, y = 0 is a horizontal asymptote.)

• The limit

  Think: Way to the right, the y-values on the graph are getting very close to the line        y = 1(so that way to the right, y = 1 is a horizontal asymptote.)

• The limit

  Think: Just to the right of x = 4, the y-values on the graph are very large and negative (this should be a vertical asymptote, then..

• The limit

  Think: Just to the left of x = 4, the y-values on the graph are very large and positive (this confirms that x = 1 is an asympote, with the graph way up just to the left of x = 1, and way down just to the right of x = 1.)

One final note:

A good way to put all of this information together is to draw any kind of graph, then erase and re-draw to make your graph fit the criteria. Anyways, my graph would look something like

True or False: If the limit

,

then f(2) = 3.

Solution:

False! The left-hand side tells you the hole is 3 units high, the right-hand side tells you the dot is 3 units high, and as we've seen, it's easy to draw a graph where the hole is 3 units high, but the the dot is at a different height.

Solution: