Chapter 1 Solutions + Summer Homework

Welcome to my math blog! If you need help with any homework or classwork problems, or would like additional explanation for a concept we've covered in class, please email me! I'll write back to you, and I'll post similar explanations / support / help here on the blog for everyone to read.

(Don't worry, I'll never post your name or any other identifying information. I just think that all your questions deserve answers that can benefit everyone in our class!)

Let's do some math!

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These are explanations that I had emailed to various students while we were working on the summer homework. Use this information as you review old problems, or while you're studying for the final exam.

Remember, my Google Drive folder contains copies of worksheets and ANSWERS (on the last page of each file). CLICK HERE TO ACCESS.

Chp 1 Review (pg. 33):

R4b) A definite integral is the area of a graph in between the x-axis and the curve. We have learned multiple ways to calculate this area: by counting squares and by finding trapezoidal areas.

To do this by counting squares, start by actually counting up all the squares! Then, you have to figure out how much each square represents. In R4b, it looks like the width is 0.5 and the height is 1 for a single square. So, each square represents 0.5 units^2. Multiply the total number of squares you counted by 0.5 to find the actual area, which is the definite integral.

To do this with the trapezoidal rule, you can find the area of each trapezoid that makes up the area, then add them up. Although it’s great to know how to use the TRAPRULE program, you’ll also need to know how to do this by hand for the test. Remember to try using the trapezoid area formula: 1/2*h(b1 + b2)

R5b) To find the limit as x approaches 3, we can’t just plug in 3 for x, because we’d end up with 0/0. Whenever this happens, try to simplify the equation first, THEN plug in 3 for x.

For this problem, the (x-3) can cancel from the top and bottom. Then, plug in 3 for x. This should give you the limit!

T answers) In the Google Drive folder here:

https://drive.google.com/file/d/0BzkKKIr5AMbQYzVpQk1yN3NfU28/view?usp=sharing

Sec 1.5 (pg. 28):

13a) In the paragraph above part “a”, it’s basically explaining that there’s a hole in the graph at x = 2 (called a “removable discontinuity”), because you can’t plug x = 2 into the equation as is. The limit answer to this problem will tell you the coordinates of the hole in the graph (2, ??).

In part “a”, it just wants you to plug in x = 2 and show that it doesn’t work. Because we are thinking about what happens when x approaches 2, but we can’t actually plug in x = 2, we could instead just plug in something really close to 2 (like x = 1.999 and x = 2.001) to find our best guess at what the limit is.

Sec 1.4 (pg. 21):
 

3 & 13 big idea) You don’t need an equation here, because they’ve given you the data points necessary to find the “bases” of each trapezoid. Imagine what the points from the table would look like if you graphed them. You’d be able to draw trapezoids from the points down to the x-axis, and find the height and bases on each trapezoid. If it helps, make your own graph with those points, then draw in the trapezoids.

3) You can’t use the TRAPRULE calculator program on this one, because they didn’t give you an equation. Think about what this graph would look like if you graphed all the points (and actually, it might be helpful for you to actually draw this out as you’re figuring it out). The x-axis is “sec” and the y-axis is “ft/sec”. You’d have a point at (0.0, 300), (0.6, 230), etc.

The bases of trapezoids are always the parallel sides, which will be the y-values in these types of problems. The height of the trapezoid is the distance between the x-values. Here, the height of each is 0.6. Then, the bases for the first trapezoid will be 300 and 230. The bases of the second trapezoid will be 230 and 150. Then 150 and 90, and so on.

Your work will look like this. Remember, trapezoid area = 1/2 * h * (b1 + b2):

1/2 * 0.6 * (300 + 230) + 1/2 * 0.6 * (230 + 150) + 1/2 * 0.6 * (150 + 90) + 1/2 * 0.6 * (90 + 40) + 1/2 * 0.6 * (40 + 0)

A quicker way to write all of this out is to factor out the common factors. All of these terms have 1/2 * 0.6, so we can factor that to the front:

1/2 * 0.6 (300 + 230 + 230 + 150 + 150 + 90 + 90 + 40 + 40 + 0)

Then, you’ll notice the only bases that show up once are the bases on the end: 300 and 0. The bases in the middle all show up twice. So, you could write it this way, where you only list the inside bases one time, but you multiply them by 2:

1/2 * 0.6 (300 + 2(230 + 150 + 90 + 40) + 0)
Obviously you don't need to write "+0", but I kept it in there for clarification.

Summer homework:

1d) This problem could have a few different answers, all of which are equivalent, but whose forms may appear different. Here’s how I arrived at the answer in the answer key.

Negative exponents mean that you can move that quantity to the denominator, and make it a positive exponent. That doesn’t mean it’ll move to the denominator of the entire problem! It just creates some reciprocals of x:

(9 - 1/x^2) / (3 + 1/x)

Next, we have quantities in both the numerator and denominator of this problem that need to be combined, but are fractions. We need a common denominator. Multiply the 9 by x^2 on top and bottom. Multiply the 3 by x on top and bottom:

(9x^2/x^2 - 1/x^2) / (3x/x + 1/x)

Now you can add the fractions in the numerator, and add the fractions in the denominator.

Numerator is now: (9x^2 - 1) / x^2

Denominator is now: (3x + 1) / x

You now have a big fraction on top divided by a big fraction on the bottom. Instead of diving by the bottom fraction, change it to “multiplying by its reciprocal”:

(9x^2 - 1) / x^2    *    x / (3x + 1)

One of the x’s on bottom can cancel with the plain x on top:

(9x^2 - 1) / x        *    1 / (3x + 1)

Then, the (9x^2 - 1) can be factored to (3x + 1)(3x - 1). This can now cancel with the (3x + 1) on the bottom, leaving you with:

(3x - 1) / x        *    1 / 1

Which simplifies to the answer you’re looking for.

11c) First, find the midpoint of the line segment from (-1, 4) to (3, 2). To find a midpoint, you find the average of the x-coordinates, and the average of the y-coordinates:

-1+3  = 1
   2

4+2  = 3
  2

So, the midpoint is (1, 3). This means that you need to find the line from (2, 3) to (1, 3). Because both y-coordinates are 3, this means the line is horizontal and is y = 3.

13c) The domain and the vertical asymptote(s) are whatever makes the denominator equal zero. Here, -1/2 is a zero of the denominator. So, the vertical asymptote is at x = -1/2 and the domain is all x’s except for -1/2.

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The general rules for end behavior are:

Degree on top is smaller:
End behavior is a horizontal asymptote at y = 0.

Degree on top is same as degree on bottom:
End behavior is a horizontal asymptote at y = top coefficient/bottom coefficient.

Degree on top is bigger:
You need to do synthetic division or long polynomial division to find the “end behavior asymptote”, which is the kind that isn’t horizontal.

Here, the degree on top and bottom is the same, so our end behavior is: y = 5/2

13di) The degree on top is bigger, so you need to use synthetic division to find the end behavior asymptote. To figure out what numbers to use in your division, factor the denominator:

(x^3 - 9) / (x - 3)(x - 4)

The zeros of the denominator are 3 and 4. Make sure to use a “zero” placeholder” for each term in your synthetic division:

Set up:
3 | 1   0   0 -9
__________


All steps:
3 | 1   0   0 -9
__ 3 _9 _27
1 3 9 18 (don’t worry about the remainder when finding end behavior)

Set up:
4 | 1   3   9
______


All steps:
4 | 1   3   9
__4 _28
1 7 37

Your end behavior asymptote is x + 7 (ignore the remainder here too).

Your vertical asymptotes are the zeros of the denominator, so they are x = 3 and x = 4.

13dii) Start by factoring top and bottom to see if anything cancels:

(x + 3)(x - 2) / (x + 2)(x - 1)

Nothing cancels, but this factoring helps, because you can now see the zeros of the denominator, which gives you the vertical asymptotes: x = -2 and x = 1.

The degree of the top and bottom is the same, so you find the end behavior asymptote by dividing the coefficient of the top by the coefficient of the bottom: y = 1/1, which is y = 1.

23cd) Both of these problems can be tackled by creating a common denominator. For 23c, multiply the first term by sqrt(y + 2) on top and bottom. This gives you:

3(y+2) / sqrt(y+2)    +    (1-2y) / sqrt(y+2)

Now you can distribute and combine like terms in the numerator:

(3y + 6 + 1 - 2y) / sqrt(y+2)

(y + 7) / sqrt(y+2)

For 23d, the 3x^-1/2 term is the same as having sqrt(x) in the denominator. We need a common denominator, so multiply the first term by 3 on both top and bottom, and multiply the second term by sqrt(x) on both top and bottom:

9 / 3sqrt(x)    +    x / 3sqrt(x)

Now you can combine like terms in the numerator:

(9 + x) / 3sqrt(x)
 

23g) First, you need to factor out any common factors in the numerator. The numerator is two giant terms, so we should look for anything shared between those two terms. It looks like their common factors are “3", an “x", a "y^2", and "cos x”. So the common factor is 3xy^2*cos x. Here’s what it looks like factored out:

Started with:   [ 3x^2y^2*cos x - 6xy^3*cos x ] / x^6y

Factored out the GCF:   [ 3xy^2*cos x (x - 2y) ] / x^6y

Now you can cancel terms from the top and bottom, since you have that GCF in front. Look for anything that’s the same on top and bottom. Here, you’ll be able to cancel an “x” and a “y”.

Canceling:   [ 3y*cos x (x - 2y) ] / x^5