Summer AP Cal Work

[CheyenneH]

Hello, I'm back again. Thanks for the help with the last problems! You guys have definitely been a huge help to me thus far, but I still have some more problems and well little time left before school. So, if you could again lend me a hand that would be great. Some of the questions I have been able to get through okay, but I'm unsure of the answer, while others are complete a total calculator problems that I'm having major difficulty with. I'm not calculator savvy at all (and I just checked I have a TI-84 not 83), so those problems on the packet I've been having major trouble with.

So if you could help me with three more problems that would be great!

1) Let \(\displaystyle \ f(x) \ = \sqrt {x-3}\) and \(\displaystyle \ g(x) \ = \ x^{2}+1\). Compute \(\displaystyle (g \ o \ f)(x)\), state its domain in interval notation.

Here is what I have so far:
\(\displaystyle (( \sqrt {x-3}^{2}+1)( \sqrt{x-3})\)
\(\displaystyle ((x-3)+1)( \sqrt {x-3})\)
\(\displaystyle (x-2)( \sqrt {x-3})\)

The domain, I have little trouble with it is the interval notation I'm not understanding. If someone could explain it to me or lead me in the right direction that would be great.

2) Find the points of intersection of:

\(\displaystyle x^{2} + y^{2} = 4\) which I get to \(\displaystyle \frac {x^{2}}{4} + \frac {y^{2}}{4}\) which I believe is the equation of a circle (have yet to graph).
&
\(\displaystyle x^{2} + y^{2} - 4x - 4y = -4\), for this one I think you do \(\displaystyle x^{2} + y^{2} - 4(x+y) = -4\), but it doesn't look right at all because from there I would divide the -4 on both sides to equal it to 1.

So, could you help me boil down the second equation to the proper equation that I can graph to find the intersections.

And lastly,

3) Given that \(\displaystyle f(x) = \frac{2x^{2}}{5x^{2}-9x-2}\) Find lim ( underneath it is a x-->? ) f(x). Also state the domain of their function.

This one I have absolutely no idea how to begin, so if you could please help me with the first step that would be great.

Thanks yet again!
-Cheyenne

 

[Aladdin]

#3) limit of f(x) as x tends to infinity is :2/5 Y=2/5 horizantol asymptote

Domain : Denomenator must be different from zero : Actually the valus of x are verticle asymptote .

 

[Deleted member 4993]

Hello, I'm back again. Thanks for the help with the last problems! You guys have definitely been a huge help to me thus far, but I still have some more problems and well little time left before school. So, if you could again lend me a hand that would be great. Some of the questions I have been able to get through okay, but I'm unsure of the answer, while others are complete a total calculator problems that I'm having major difficulty with. I'm not calculator savvy at all (and I just checked I have a TI-84 not 83), so those problems on the packet I've been having major trouble with.

So if you could help me with three more problems that would be great!

1) Let \(\displaystyle \ f(x) \ = \sqrt {x-3}\) and \(\displaystyle \ g(x) \ = \ x^{2}+1\). Compute \(\displaystyle (g \ o \ f)(x)\), state its domain in interval notation.

Every book has definition of (f o g) and those are not equivalent - almost!! In your textbook - how is that operation defined? In general,

(g o f)(x) = g[f(x)]

If you are using the same definition - then you should not multiply with f(x)

(g o f)(x) = g[?(x-3)] = [?(x-3)][sup:limgal63]2[/sup:limgal63] + 1 = x - 2

however, watch out for domain. f(x) has restricted domain [3, ?)

go to

http://www.analyzemath.com/CompositionF ... rials.html

for a quick review

Here is what I have so far:
\(\displaystyle (( \sqrt {x-3}^{2}+1)( \sqrt{x-3})\)
\(\displaystyle ((x-3)+1)( \sqrt {x-3})\)
\(\displaystyle (x-2)( \sqrt {x-3})\)

The domain, I have little trouble with it is the interval notation I'm not understanding. If someone could explain it to me or lead me in the right direction that would be great.

2) Find the points of intersection of:

\(\displaystyle x^{2} + y^{2} = 4\) which I get to \(\displaystyle \frac {x^{2}}{4} + \frac {y^{2}}{4}\) which I believe is the equation of a circle (have yet to graph).
&
\(\displaystyle x^{2} + y^{2} - 4x - 4y = -4\), for this one I think you do \(\displaystyle x^{2} + y^{2} - 4(x+y) = -4\), but it doesn't look right at all because from there I would divide the -4 on both sides to equal it to 1.

So, could you help me boil down the second equation to the proper equation that I can graph to find the intersections.

x[sup:limgal63]2[/sup:limgal63] + y[sup:limgal63]2[/sup:limgal63] - 4x - 4y = -4

Complete square

x[sup:limgal63]2[/sup:limgal63] - 4x + 4 + y[sup:limgal63]2[/sup:limgal63] - 4y + 4 = -4 + 4 + 4

(x - 2)[sup:limgal63]2[/sup:limgal63] + (y - 2)[sup:limgal63]2[/sup:limgal63] = 2[sup:limgal63]2[/sup:limgal63]

Look familiar??!!




And lastly,

3) Given that \(\displaystyle f(x) = \frac{2x^{2}}{5x^{2}-9x-2}\) Find lim ( underneath it is a x-->? ) f(x). Also state the domain of their function.

Divide numerator and the denominator by x[sup:limgal63]2[/sup:limgal63] - then take limit

This one I have absolutely no idea how to begin, so if you could please help me with the first step that would be great.

Thanks yet again!
-Cheyenne

 

[CheyenneH]

Okay thanks with the first and second one, now the only one I am having trouble with is #3. Mainly because I have never worked with limits before, so would there be a way to do it on the calculator to check it because the teacher wants us to be familiar with our calculators or should I figure it out without one?

If without, would I:

3) \(\displaystyle \frac {2x^{2}}{5x^{2}-9x-2} * \frac {x^{2}}{x^{2}}\)

Or:

\(\displaystyle \frac {\frac{2x^{2}}{x^{2}}}{\frac{5x^{2}-9x-2}{x^{2}}}\)

 

[mmm4444bot]

… would there be a way to do it on the calculator to check it …

Hello Cheyenne:

Of course, nobody can determine whether or not this particular calculator can be used to "check" a limit, until after you tell us the manufacturer and model number. (Do you have the user's manual?)

'

\(\displaystyle \frac {2x^{2}}{5x^{2}-9x-2} * \frac {x^{2}}{x^{2}}\)

NO - this is not dividing numerator and denominator by x^2; instead, this is multiplying by x^2.

Or:

\(\displaystyle \frac {\frac{2x^{2}}{x^{2}}}{\frac{5x^{2}-9x-2}{x^{2}}}\)YES - now do the divisions (i.e., simplify by canceling common factors).

After you simplify, you'll have some rational terms in which the numerator is constant while the denominator is growing without bound (because it contains a power of x, and we're letting x increase forever, in this limit).

What happens to the value of a ratio when the numerator is fixed and the denominator keeps growing?

1/x approaches ???? as x approaches infinity

Cheers,

~ Mark

 

[Deleted member 4993]

Mark already explained to you the steps that need to be taken - I'll just take few more steps to complete the problem
3)

\(\displaystyle \frac {\frac{2x^{2}}{x^{2}}}{\frac{5x^{2}-9x-2}{x^{2}}}\)

\(\displaystyle = \, \frac {2}{5 - \frac{9}{x} - \frac{2}{x^2}}\)

as x becomes very very very large - (9/x) and (2/x[sup:1a4jigua]2[/sup:1a4jigua]) become very very small almost equal to zero (compared to 5)

So as x ? ? we have f(x) ? 2/5

 

[Aladdin]

Mark already explained to you the steps that need to be taken - I'll just take few more steps to complete the problem
3)

\(\displaystyle \frac {\frac{2x^{2}}{x^{2}}}{\frac{5x^{2}-9x-2}{x^{2}}}\)

\(\displaystyle = \, \frac {2}{5 - \frac{9}{x} - \frac{2}{x^2}}\)

as x becomes very very very large - (9/x) and (2/x[sup:ktndceul]2[/sup:ktndceul]) become very very small almost equal to zero (compared to 5)

So as x ? ? we have f(x) ? 2/5

Another way - You can choose the largest power of the numerator and the denomenator and simplify - Easier.

 

[stapel]

Another way - You can choose the largest power of the numerator and the denomenator and simplify - Easier.

They are variable expressions. What are "the largest values" of each? And on what basis and by what method are you applying these values to the process of determining the limit? :?

 

[mmm4444bot]

Another way - You can choose the largest power of the numerator and the denomenator and simplify - Easier

I'm not sure what you're trying to say, Aladdin.

Using the largest power of x to simplify the limit calculation has already been done. Can you explain in more detail what you're thinking?

An "easier" method might be to consider the rules for determining the horizontal asymptote of a rational function.

If the degrees of the numerator and denominator are equal (as they are in this exercise), then the horizontal asymptote (i.e., the limit as x approaches infinity) is simply the ratio of the leading coefficients (2/5).

I think it's a good idea for people to understand the basis behind alternate methods, before using them. The basis for taking the ratio of leading coefficients, in this exercise, can be demonstrated (in the general case) by the division that Subhotosh first described.

Cheers,

~ Mark

 

[Aladdin]

Example : The largest power in the numerator was 2 ... and in the denomenator was also 2 .

So I choosed from the numerator 2x^2 and from the denomenator 5x^2 --

Which leads me to ( offcourse as x tends to infinity) : \(\displaystyle \frac{2x^2}{5x^2}\)

Simplify :

\(\displaystyle \frac{2}{5}\)

 

[mmm4444bot]

So I choosed from the numerator 2x^2 and from the denomenator 5x^2

This method only works under certain conditions (just like that quadratic-formula variation thingy that you posted earlier this year).

What is the basis for this easier method? In other words, how does your reasoning above differ from what's already been posted about this particular limit?

 

[Aladdin]

Re:

So I choosed from the numerator 2x^2 and from the denomenator 5x^2

This method only works under certain conditions (just like that quadratic-formula variation thingy that you posted earlier this year).

What is the basis for this easier method? In other words, how does your reasoning above differ from what's already been posted about this particular limit?

Yes , under particular conditions mark . -- Why would I have to devide by x^2 the numerator and the denomanator & I can know it faster.

 

[mmm4444bot]

Why would I have to devide by x^2 the numerator and the denomanator & I can know it faster.

I think that I already answered this question, but, if you take time to answer my questions, then I will take time to answer yours.

 

[Deleted member 4993]

[
Another way - You can choose the largest power of the numerator and the denomenator and simplify - Easier.

The short-cut above only works when the orders of the polynomials (in the numerator and the denominators) are equal.

It is very dangerous to discuss short-cuts while you are teaching at the elementary levels of problem-solving.

 

[Aladdin]

[
Another way - You can choose the largest power of the numerator and the denomenator and simplify - Easier.

The short-cut above only works when the orders of the polynomials (in the numerator and the denominators) are equal.

It is very dangerous to discuss short-cuts while you are teaching at the elementary levels of problem-solving.

\(\displaystyle \frac{2x^3+2}{2x^2}\) as x tends to infinity

=\(\displaystyle \frac{2x^3}{2x^2}\)

= \(\displaystyle x\)

=\(\displaystyle \infty\)

= infinity