# Help me with my homework

I thought such a title would draw people in to my topic.

Okay. I got a large calculus test coming up on Tuesday. I am currently reviewing like mad. I’ve worked out some of these questions, and my instructor said they were done incorrectly. There’s about 100 review questions, but I’ll pick out like 10, one from each time, and put it on these here forums. If you all can get the answer, with proving it with steps, and post it here; it’d help me critically in my understanding of calculus.

<strike><b>1. In a geometric sequence of real numbers, the sum of the first six terms is nine times the sum of the first three terms. If the first term is 5, what is the third term?</b></strike>

<strike><b>2. Let f(x) = {x^3 - 6x^2 + 9x, for 0<&=x, and x<&=4},{-x + 7, for 4<x, and x<&=7}

a) For 0<&=x, and x<&=7, determine the intervals in which the function <i>f</i> is decreasing.
b) Ommitted.
c) Does Lim x->∞ exist? Explain.</strike>

<strike>3. The sum of the infinite series, c/2 + (c/2)^2 + (c/2)^3 + … is 2. Determine the value of <i>c</i>.</strike>

<strike>4. Consider the arithmetic sequence where t1 + t3 + t5 + t7 + t9 = 17, and t2 + t4 + t6 + t8 + t10 = 15. Find <i>d</i>.</b></strike>

<b>5. Find the arithmetic sequence with first term = 1, and constant difference is not 0, whose second, tenth, and thirty-fourth terms are the first three terms on a geometric sequence.</b>

Yeah, just reading that confuses me.

<strike><b>6. At what numbers is F(x) = (3|x|-x^2)/([x^0.5]+x^(1/3))</b></strike>

There’s like a billion others, but I got those. But if you can explain with numbers and/or words how you answer them or even show the answers, it’d be REALLY appreciated.

ANSWERED: 1, 2, 3, 4, 6

To question 1:

Formula’s needed, Tn = ar^[n-1], and Sn = a[[r^n]-1]/[r-1]

T3 = needed
T1 = 5

a = 5

S6 = 9*S3, so we use the Sn formula.

5*[[r^6]-1]/[r-1] = 95[[r^3]-1]/[r-1]

Divide both by r-1, and multiply the 9 and five, you end up with:

5*[[r^6]-1] = 45*[[r^3]-1]

Expand.

5r^6 - 5 = 45r^3 - 45

Bring it over.

5r^6 - 45r^3 + 40 = 0

Divide it by 5.

r^6 - 9r^3 + 8 = 0

I randomly try numbers, and I put in 2 for r.

2^6 - 9[2]^3 + 8 = 0

Therefore, 2 is r.

a = 5
r = 2
n = 3, since I’m trying to find the third term.

T3 = [5][2]^[3-1]
T3 = [5][2]^[2]
T3 = [5][4]
T3 = 20

Hooray!

Number 67 has to do with geometric series. If you have a series of the form

S = a + ar + a(r^2) + a*(r^3) + …

then its sum is

S = a/(1-r).

In this case, a = r = (c/2) and S = 2, so you have

2 = (c/2) / (1 - (c/2) ).

You can easily solve for c.

Bring back the homework board!

Actually, I’ve been sitting here for like 5 minutes trying to find out how to solve for c. Explain.

edit;
Oh shit, nevermind.

(c/2) / (1-(c/2)) = 2
2(c/2) / 2(1-(c/2)) = 2
c / 2 - c = 2
c = [2 - c] *2
c = 4 - 2c
3c = 4
c = 1.333333333333333333333333333333333333333333

Correct?

OMG… Im not going to advanced math next year. I quit if I would eventually have to pull THAT off!!!

Maybe you could write the formula of Question 2 so that it actually makes sense. For example, “4<&=7” doesn’t make sense, since 4 is always < 7 and never = 7.
Edit: I tried to figure out what you mean, and it’s decreasing from ]1;3[ and ]4;7[
To determine when this first long term is decreasing, you have to derive it; the derived function f’(x) is f’(x)=3x^2 - 12x + 9; now, find out where the derived function is zero (i.e. where the slope is zero; those are typically points where the function changes its monoty). Those points are 1 and 3. Now check whether it’s really decreasing in that interval by using two samples. You’ll see that it’s really decreasing there, so you have your first interval. The second is easy, because in ]4;7] the function is a linear function with negative slope.
For c), it doesn’t exist because the function dies after 7.

I apologize. But I meant 4<&=x, x<&=7.

Edit; derive? Is there a way to answer the question without derivatives?

I’m sure you have learnt strategies for tasks like that. How have you solved stuff like that in class? I mean, it’s possible that there is a way to do it without derivatives…

<img src = “http://www.rpgclassics.com/staff/cless/mathhelpsm.gif”>

Number 31 works on roughly the same logic as number 42, so I’ll give you some time to work it out.

Also I don’t know what number 6 is until you understand what nth roots are >:E

If you don’t understand something, I’ll try to clarify, but if you don’t know what derivatives are then there is something wrong with your calculus course.

edit: also for question 1, I ignored the “-x+7” part because it didn’t make sense A function cannot be both x^3 + 6x^2 + 9x and -x + 7 at the same x-coordinate.

Thanks Cless.

The function contains two different graphs. Let’s say you had a function of f(x) = {x^2}, {x}. That means that there’s a parabola, and a diagonal line. I would draw it, but I don’t got a scanner. Just think of it as two graphs put together. A parabola with a line going diagonal from whatever to whatever. It’s difficult to explain.

And, we just began the calculus course. We’ve only reviewed for geometric & arithmetic sequences and series, and pretty much did how to solve limits; and we’ve touched a little on continuity. Derivatives are next unit, which is after this upcoming test.

edit;

1. Let f(x) = {x^3 - 6x^2 + 9x, for 0<&=x, and x<&=4},{-x + 7, for }.

What I meant by 0<&=x, and x<&=4, and 4<&=x and x<&=7, is that “<&=” means less than or equal. I can’t find a symbol for that bracket thing with an underline. Unless I did <u><</u>.

I knew what you mean by <&=, but it’s not possible for a function to be described by two different operations of x at the same x-coordinate. One of the basic properties that defines a <i>function</i> is that for any given x-coordinate, there is one, and only one, y-value cooresponding with it. If there are more than one, then it is not a function (of x, anyway).

Either your f(x) is x^2 on some interval and x on a different interval (these are called piecewise functions), or f(x) is x^2 and g(x) is x.

If they’re two different functions, then g(x) is decreasing for all x. (the derivative of -x+7 is -1)
If it’s a piecewise function, then whenever f(x) = -x + 7 doesn’t even have to be mentioned because it’s always decreasing.

Another way of solving number 2 is that the expression can be factored to x(x-3)^2. Because the entire thing is multiplied by x, and (x-3)^2 is never negative, the interval from 0 to some number must be increasing. You can see that this “some number” is 1 because at that point, (1-3) is squared. (if it was 2, (2-3) squared would only get you 1).

The function (x-3)^2 is simply increasing from x=3 to x=inf., so when multiplied again by x, whenever x is a positive number above 3 the function must be increasing. Because the upper bound is 4, the function is increasing from 3 to 4.

It’s a piecewise (two part) function.

The teacher drew it as f(x) then a { but inside that { are two different equations. Then she closed it with }.

Okay, then I think you mean it is -x+7 on the interval x>4, or something else?

The function, once graphed, shows the first function when 0 <u><</u> x <u><</u> 4, and the second function when 4 < x <u><</u> 7.

Ohhhhhhhhhhhh okay. Then you don’t need to worry about the second part at all (since it’s always decreasing), except say in 2c) that it is bound on the right by 7 instead of 4.

edit: yeah this is much clearer than the <&= notation

When it asks for when the function is decreasing, is it when y is decreasing, or when x is decreasing?

edit; in your explanation, you said that the function increases and decreases at…

0,1 = inc
1,3 = dec
3,4 = inc

I am graphing it on my graphic calculator, and neither piece of the function touches any of those points.

I get;

0,0 = inc
1,4 = dec
3,0 = inc
4,3.5 = dec

y is another way of writing f(x), which is saying “a function of x”.

edit: It’s easier to understand the concept of functions if you think of f( ) as a transformation machine.

For example, f(x) = x^2 describes how the transformation machine transforms any given x by squaring it. z(x) = 4 - x describes how the transformation machine transforms any given x by subtracting it from 4. g(x) = s would be a <i>constant</i> value s because the machine says with g(<b>x</b>) that it will transform x, not s, even though s is not a number (the specific example says that it will <i>transform anything put into the function by turning it into the value <b>s</b></i>). Same deal if you had h(s) = x; you’d get the same value, <i>x</i>, out each time, because the machine says it will transform s.

To solidify the above examples:

f(x) = x^2
x = 2; f(2) = 4
x = 6; f(6) = 36

g(x) = s
x = 6; g(6) = s
x = 26326236; g(26326236) = s
x = beard of abraham lincoln; g(beard of abraham lincoln) = s

h(s) = x
s = 64; h(64) = x
s = x; h(x) = x
s = -2; h(-2) = x

edit: f( ), g( ), h( ), z( ) are just different labels for different functions to make this example clearer

I editted my above post, so check it oot.

Okay, I have it all down, except two of the questions.

5 and 6. I renumbered the questions to make it look more logical.

By using this, I was able to graph number 6. It shows that the function never reaches x=0, or y=0, nor below either of those. So the function is discontinuous at (0,0), since there is a (1,1) if you make tables representing y and x vaues. The lowest x value is 1, the highest x value is supposedly infinity. The lowest y-value is infinity supposedly as well, and the highest is 1. The graph goes up until 1,1, then declines with 2,0.7, then 3,0, then 4,-1.1, and so on.