precalculus

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course mth 152

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Question: `q001 A straight line connects the points (3, 5) and (7, 17),

while another straight line continues on from (7, 17) to the point (10,

29). Which line is steeper and on what basis to you claim your result?

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Your solution:

The line from (7,17) to (10,29) is steeper because there is a 3 point difference along the x axis as opposed to the longer distance of 4 points for (3,5) to (7,17). This coupled with the difference of 17 to 29 on the y axis makes in significantly steeper.

confidence rating #$&*:1

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Given Solution:

`aThe point (3,5) has x coordinate 3 and y coordinate 5. The point (7,

17) has x coordinate 7 and y coordinate 17. To move from (3,5) to (7,

17) we must therefore move 4 units in the x direction and 12 units in

the y direction.

Thus between (3,5) and (7,17) the rise is 12 and the run is 4, so the

rise/run ratio is 12/4 = 3.

Between (7,10) and (10,29) the rise is also 12 but the run is only 3--

same rise for less run, therefore more slope. The rise/run ratio here

is 12/3 = 4.

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Self-critique (if necessary):

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Question: `q002. The expression (x-2) * (2x+5) is zero when x = 2 and

when x = -2.5. Without using a calculator verify this, and explain why

these two values of x, and only these two values of x, can make the

expression zero.

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Your solution:

When x=2 the left side of the equation equals zero, with our multiplication including a zero our answer is zero. When x=-2.5 is the right side of the equation equaling zero. The two differing x values will change either side of the equation leaving us at zero

confidence rating #$&*:3

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Given Solution:

`aIf x = 2 then x-2 = 2 - 2 = 0, which makes the product (x -2) * (2x +

5) zero.

If x = -2.5 then 2x + 5 = 2 (-2.5) + 5 = -5 + 5 = 0.which makes the

product (x -2) * (2x + 5) zero.

The only way to product (x-2)(2x+5) can be zero is if either (x -2) or

(2x + 5) is zero.

Note that (x-2)(2x+5) can be expanded using the Distributive Law to get

x(2x+5) - 2(2x+5). Then again using the distributive law we get

2x^2 + 5x - 4x - 10 which simplifies to

2x^2 + x - 10.

However this doesn't help us find the x values which make the

expression zero. We are better off to look at the factored form.

STUDENT QUESTION

I think I have the basic understanding of how x=2 and x=-2.5 makes this

equation 0

I was looking at the distributive law and I understand the basic

distributive property as stated in algebra

a (b + c) = ab + ac and a (b-c) = ab - ac

but I don’t understand the way it is used here

(x-2)(2x+5)

x(2x+5) - 2(2x+5)

2x^2 + 5x - 4x - 10

2x^2 + x - 10.

Would you mind explaining the steps to me?

INSTRUCTOR RESPONSE

The distributive law of multiplication over addition states that

a (b + c) = ab + ac

and also that

(a + b) * c = a c + b c.

So the distributive law has two forms.

In terms of the second form it should be clear that, for example

(x - 2) * c = x * c - 2 * c.

Now if c = 2 x + 5 this reads

(x-2)(2x+5) = x * ( 2 x + 5) - 2 * (2 x + 5).

The rest should be obvious.

We could also have used the first form.

a ( b + c) = ab + ac so, letting a stand for (x - 2), we have

(x-2)(2x+5) = ( x - 2 ) * 2x + (x - 2) * 5.

This will ultimately give the same result as the previous. Either way

we end up with 2 x^2 + x - 10.

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Self-critique (if necessary):

Looking at the above examples the distributive law didn't cross my mind, I just worked out the equations.

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Question: `q003. For what x values will the expression (3x - 6) * (x +

4) * (x^2 - 4) be zero?

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Your solution:

??? at a bit of a loss

confidence rating #$&*:0

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Given Solution:

`aIn order for the expression to be zero we must have 3x-6 = 0 or x+4=0

or x^2-4=0.

3x-6 = 0 is rearranged to 3x = 6 then to x = 6 / 3 = 2. So when x=2,

3x-6 = 0 and the entire product (3x - 6) * (x + 4) * (x^2 - 4) must be

zero.

x+4 = 0 gives us x = -4. So when x=-4, x+4 = 0 and the entire product

(3x - 6) * (x + 4) * (x^2 - 4) must be zero.

x^2-4 = 0 is rearranged to x^2 = 4 which has solutions x = + - `sqrt(4)

or + - 2. So when x=2 or when x = -2, x^2 - 4 = 0 and the entire

product (3x - 6) * (x + 4) * (x^2 - 4) must be zero.

We therefore see that (3x - 6) * (x + 4) * (x^2 - 4) = 0 when x = 2, or

-4, or -2. These are the only values of x which can yield zero.**

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Self-critique (if necessary):

I should have understood that, I made it more complicated than it had to be.

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Question: `q004. One straight line segment connects the points (3,5)

and (7,9) while another connects the points (10,2) and (50,4). From

each of the four points a line segment is drawn directly down to the x

axis, forming two trapezoids. Which trapezoid has the greater area? Try

to justify your answer with something more precise than, for example,

'from a sketch I can see that this one is much bigger so it must have

the greater area'.

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Your solution:

In my graph the first two points (3,5) and (7,9)reacher higher on the y axis the points (10,2) and (50,4) have much higher area covered on the x axis rages from 10 to 50 giving it a much greater surface area.

confidence rating #$&*:1

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Given Solution:

`aYour sketch should show that while the first trapezoid averages a

little more than double the altitude of the second, the second is

clearly much more than twice as wide and hence has the greater area.

To justify this a little more precisely, the first trapezoid, which

runs from x = 3 to x = 7, is 4 units wide while the second runs from x

= 10 and to x = 50 and hence has a width of 40 units. The altitudes of

the first trapezoid are 5 and 9,so the average altitude of the first is

7. The average altitude of the second is the average of the altitudes 2

and 4, or 3. So the first trapezoid is over twice as high, on the

average, as the first. However the second is 10 times as wide, so the

second trapezoid must have the greater area.

This is all the reasoning we need to answer the question. We could of

course multiply average altitude by width for each trapezoid, obtaining

area 7 * 4 = 28 for the first and 3 * 40 = 120 for the second. However

if all we need to know is which trapezoid has a greater area, we need

not bother with this step.

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Question: `q005. Sketch graphs of y = x^2, y = 1/x and y = `sqrt(x)

[note: `sqrt(x) means 'the square root of x'] for x > 0. We say that a

graph increases if it gets higher as we move toward the right, and if a

graph is increasing it has a positive slope. Explain which of the

following descriptions is correct for each graph:

As we move from left to right the graph increases as its slope

increases.

As we move from left to right the graph decreases as its slope

increases.

As we move from left to right the graph increases as its slope

decreases.

As we move from left to right the graph decreases as its slope

decreases.

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Your solution:

1,1.415,1.732,2. The graph increases but at a decreasing rate.

confidence rating #$&*:3

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Given Solution:

`aFor x = 1, 2, 3, 4:

The function y = x^2 takes values 1, 4, 9 and 16, increasing more and

more for each unit increase in x. This graph therefore increases, as

you say, but at an increasing rate.

The function y = 1/x takes values 1, 1/2, 1/3 and 1/4, with decimal

equivalents 1, .5, .33..., and .25. These values are decreasing, but

less and less each time. The decreasing values ensure that the slopes

are negative. However, the more gradual the decrease the closer the

slope is to zero. The slopes are therefore negative numbers which

approach zero.

Negative numbers which approach zero are increasing. So the slopes are

increasing, and we say that the graph decreases as the slope increases.

We could also say that the graph decreases but by less and less each

time. So the graph is decreasing at a decreasing rate.

For y = `sqrt(x) we get approximate values 1, 1.414, 1.732 and 2. This

graph increases but at a decreasing rate.

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Self-critique (if necessary):

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Question: `q006. If the population of the frogs in your frog pond

increased by 10% each month, starting with an initial population of 20

frogs, then how many frogs would you have at the end of each of the

first three months (you can count fractional frogs, even if it doesn't

appear to you to make sense)? Can you think of a strategy that would

allow you to calculate the number of frogs after 300 months (according

to this model, which probably wouldn't be valid for that long) without

having to do at least 300 calculations?

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Your solution:

1st month=22

2nd month=24.2

3rd month=26.62

The actual rate of increase is 1.1

so 20*1.1^300=a very a large number that i need a calculator for

confidence rating #$&*:3

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Given Solution:

`aAt the end of the first month, the number of frogs in the pond would

be (20 * .1) + 20 = 22 frogs. At the end of the second month there

would be (22 * .1) + 22 = 24.2 frogs while at the end of the third

month there would be (24.2 * .1) + 24.2 = 26.62 frogs.

The key to extending the strategy is to notice that multiplying a

number by .1 and adding it to the number is really the same as simply

multiplying the number by 1.1. We therefore get

20 * 1.1 = 22 frogs after the first month

22 * 1.1 = 24.2 after the second month

etc., multiplying by for 1.1 each month.

So after 300 months we will have multiplied by 1.1 a total of 300

times. This would give us 20 * 1.1^300, whatever that equals (a

calculator, which is appropriate in this situation, will easily do the

arithmetic).

A common error is to say that 300 months at 10% per month gives 3,000

percent, so there would be 30 * 20 = 600 frogs after 30 months. That

doesn't work because the 10% increase is applied to a greater number of

frogs each time. 3000% would just be applied to the initial number, so

it doesn't give a big enough answer.

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Question: `q007. Calculate 1/x for x = 1, .1, .01 and .001. Describe

the pattern you obtain. Why do we say that the values of x are

approaching zero? What numbers might we use for x to continue

approaching zero? What happens to the values of 1/x as we continue to

approach zero? What do you think the graph of y = 1/x vs. x looks for x

values between 0 and 1?

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Your solution:

x =

1=1

.1=10

.01=100

.001=1000

The numbers will increase exponentially because it can never actually reach zero. Because it can never reach zero the the graph would just keep going without it ever actually reaching zero on the y axis.

confidence rating #$&*:

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Given Solution:

`aIf x = .1, for example, 1 / x = 1 / .1 = 10 (note that .1 goes into 1

ten times, since we can count to 1 by .1, getting.1, .2, .3, .4, ...

.9, 10. This makes it clear that it takes ten .1's to make 1.

So if x = .01, 1/x = 100 Ithink again of counting to 1, this time by

.01). If x = .001 then 1/x = 1000, etc..

Note also that we cannot find a number which is equal to 1 / 0. Deceive

why this is true, try counting to 1 by 0's. You can count as long as

you want and you'll ever get anywhere.

The values of 1/x don't just increase, they increase without bound. If

we think of x approaching 0 through the values .1, .01, .001, .0001,

..., there is no limit to how big the reciprocals 10, 100, 1000, 10000

etc. can become.

The graph becomes steeper and steeper as it approaches the y axis,

continuing to do so without bound but never touching the y axis.

This is what it means to say that the y axis is a vertical asymptote

for the graph .

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Self-critique (if necessary):

My answers are very similar but I have zero confidence before looking at the solution

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Question: `q008. At clock time t the velocity of a certain automobile

is v = 3 t + 9. At velocity v its energy of motion is E = 800 v^2. What

is the energy of the automobile at clock time t = 5?

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Your solution:

v=15+9, v=24

E= 800 24^2

E= 800*576=

E=460800

confidence rating #$&*:0

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Given Solution:

`aFor t=5, v = 3 t + 9 = (3*5) + 9 = 24. Therefore E = 800 * 24^2 =

460800.

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Question: `q009. Continuing the preceding problem, can you give an

expression for E in terms of t?

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Your solution:

E =800 ( 3t + 9) ^2

just reverse worked the equation???

confidence rating #$&*:1

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Given Solution:

`aSince v = 3 t + 9 the expression would be E = 800 v^2 = 800 ( 3t + 9)

^2. This is the only answer really required here.

For further reference, though, note that this expression could also be

expanded by applying the Distributive Law:.

Since (3t + 9 ) ^ 2 = (3 t + 9 ) * ( 3 t + 9 ) = 3t ( 3t + 9 ) + 9 * (3

t + 9) = 9 t^2 + 27 t + 27 t + 81 = 9 t^2 + 54 t + 81, we get

E = 800 ( 9 t^2 + 54 t + 81) = 7200 t^2 + 43320 t + 64800 (check my

multiplication because I did that in my head, which isn't always

reliable).

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Question: For what x values is the value of the expression (2^x - 1)

( x^2 - 25 ) ( 2x + 6) zero?

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Your solution:

0,25,-3 will give us zero in each subsequent equation

confidence rating #$&*:3

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Question: One straight line segment connects the points (3,5) and

(7,9) while another connects the points (3, 10) and (7, 6). From each

of the four

points a line segment is drawn directly down to the x axis, forming

two trapezoids. Which trapezoid has the greater area?

Any solution is good, but a solution that follows from a good argument

that doesn't actually calculate the areas of the two trapezoids is

better.

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Your solution:

Both of the sets max out the x axis at 7 but the real difference is that points (3, 10) and (7, 6) do cover a much wider area one the y axis is taken into account

confidence rating #$&*:3

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Question:

Suppose you invest $1000 and, at the end of any given year, 10% is

added to the amount. How much would you have after 1, 2

and 3 years?

What is an expression for the amount you would have after 40 years

(give an expression that could easily be evaluated using a calculator,

but don't bother to actually evaluate it)?

What is an expression for the amount you would have after t years?

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Your solution:

1000*.10=1100

1100*.10=1210

1210*.10=1331

confidence rating #$&*:3

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