N help with easy calc problems

[Kulith]

I made the mistake of zoning out this week in calculus and now I have no idea what where talking about. I think these are easy problems, I just have no idea how to do them. I cant sleep and pay attention at the same time you know.

[sqrt] (x-1) = the square root of (x - 1)

Here they go:

Locate the absolute extrema of the function f(x) = 2x - 3 over the indicated interval: - say what?

a. (0,2]
b. [0,2]
c. (0,2)

~~~~~~~~~~~~~~~~~~~~~~~

Show your work to find the indicated limit

Lim x->infinity of (1-2x) / ( [sqrt](4x[sup:3vene50u]2[/sup:3vene50u]+x) )

~~~~~~~~~~~~~~~~~~~~~~~

Find the domain of the function:

a. f(x) = [sqrt](x^4 - 16x[sup:3vene50u]2[/sup:3vene50u])

b. f(x) = [sqrt](1 - sin(x))

c. f '(x) = [sqrt](1 - sin(x))

~~~~~~~~~~~~~~~~~~~~~~~

Any help would be great. I really need the points now, and im desperate enough to find a help forum...

The sheet is due monday, so if I could get help before then it would be awesome.

 

[stapel]

Locate the absolute extrema of the function f(x) = 2x - 3 over the indicated interval: - say what?

Do you need lessons on what "extrema" are, how to find them, what "absolute" extrema are, and/or how to find them? We'll be glad to try to find lesson-links to "cure" your "say what?" feeling, but you'll need to specify how far we need to go back. :idea:

Show your work to find the indicated limit

Lim x->infinity of (1-2x) / ( [sqrt](4x[sup:tn237txg]2[/sup:tn237txg]+x) )

Are you supposed to use epsilon-delta definitions, or simpler methods? :?:

(Limits generally take a few weeks to cover, so, since you only missed a week ending somewhere in derivatives, obviously you were in class for this topic. You'll need to specify what method you're using, and show what you've done and where you're stuck.)

Find the domain of the function:

You learned how to do this back in algebra: The domain is all allowable x-values, so check for any "problem" values (negatives inside square roots, zeroes in denominators, etc), and then the domain is everything else! :wink:

Eliz.

 

[galactus]

It's probably not a good idea to freely admit on the forum that you have been 'sleeping', or at least not paying attention, in class. You will garner little sympathy. :wink:

Here's a hint for the limit problem.

\(\displaystyle \lim_{x\to\infty}\frac{1-2x}{\sqrt{4x^{2}+x}}\)

Divide the numerator by x and the denominator by \(\displaystyle \sqrt{x^{2}}\)

It will then be easy to see the limit.

 

[Kulith]

Do you need lessons on what "extrema" are, how to find them, what "absolute" extrema are, and/or how to find them? We'll be glad to try to find lesson-links to "cure" your "say what?" feeling, but you'll need to specify how far we need to go back. :idea:

I know weve done this in class. I know you guys can probably just look at this problem and know the answer. I dont know why there is a difference between open and closed intervals in this problem. I know this problem is easy, i just dont remember how to do it. I remember her doing problems like this on the board, but when I go home and look at the problem my mind is blank . I just need a boost in the right direction.

Are you supposed to use epsilon-delta definitions, or simpler methods? :?:

Seeing as ive never heard of epsilon-delta in my life, Id say its simpler methods. Weve done laods of limits, I just cant remember what im supposed to do with infinity. Somehow I need to endup with 1/infinity or something like that, which would = 0.

You learned how to do this back in algebra: The domain is all allowable x-values, so check for any "problem" values (negatives inside square roots, zeroes in denominators, etc), and then the domain is everything else! :wink:

I know how to find the domain usually. Ive done all the problems that I can. But for these three, im not sure what process im supposed to do. Because there are consistent intervals of numbers that cant be in the domain, im not sure how to set up the problem. How do I find what x values of 1 - sin(x) > 0??

It's probably not a good idea to freely admit on the forum that you have been 'sleeping', or at least not paying attention, in class. You will garner little sympathy. :wink:

Here's a hint for the limit problem.

\(\displaystyle \lim_{x\to\infty}\frac{1-2x}{\sqrt{4x^{2}+x}}\)

Divide the numerator by x and the denominator by \(\displaystyle \sqrt{x^{2}}\)

It will then be easy to see the limit.

Look, this is the short story:

I have a regular calc teacher that I like very much. He explains everything in a way that I can follow along easily and learn it properly. Now I have a student teacher for a few weeks. It is extremely hard to pay attention to her as she is attempting teaching, and its hard to follow her. Not to mention she constatly makes mistakes to the point where I no longer remember which method works and which doesn't. So this past week ive found myself losing focus in her class.

That method sounds very familiar, i remember her doing a problem like that. I still dont know how to do it though.

How can I divide the numerator by x when only one of the numerator digits has the x-variable. Same with the denom....

For the denom, is \(\displaystyle \sqrt{4x^{2}}\) - \(\displaystyle \sqrt{x}\) the same as \(\displaystyle \sqrt{4x^{2} - x}\)?

 

[galactus]

\(\displaystyle \lim_{x\to\infty}\frac{1-2x}{\sqrt{4x^{2}+x}}dx\)

Divide numerator by x and denominator by \(\displaystyle |x|=\sqrt{x^{2}}\)

\(\displaystyle \lim_{x\to\infty}\frac{\frac{1}{x}-\frac{2x}{x}}{\sqrt{\frac{4x^{2}}{x^{2}}+\frac{x}{x^{2}}}}\)

You should see this simplifies to:

\(\displaystyle \lim_{x\to\infty}\frac{\frac{1}{x}-2}{\sqrt{4+\frac{1}{x}}}\)

As \(\displaystyle \rightarrow{\infty}}\), then we have \(\displaystyle \frac{-2}{\sqrt{4}}=\boxed{-1}\)

See?. If had approached negative infinity, you could have divided the numerator by -x.

 

[stapel]

I dont know why there is a difference between open and closed intervals in this problem.

Are the endpoints of an open interval included within the interval? Now think about the meaning of limits, and max/min values within that context. For instance, if you have the function g(x) = x, is there a "maximum value" on the open interval (1, 2)? Or do you only "get closer and closer" to some (unincluded) maximum value?

I just need a boost in the right direction.

Follow the process you've always followed for finding max/min points. Then, as the book's lesson and examples show, also find the values of the function at the interval endpoints (which are included in a closed interval). Compare the various values to find the biggest (this would be the maximum) and the smallest (this would be the minimum).

Weve done laods of limits, I just cant remember what im supposed to do with infinity.

Well, start by reviewing the section in your book, re-reading the explanations and following through the worked examples. It might also help to think about the horizontal asymptotes (for rational functions) that you did back in algebra. "Limits at infinity" are pretty much the same thing. :wink:

The domain is all allowable x-values, so check for any "problem" values (negatives inside square roots, zeroes in denominators, etc), and then the domain is everything else!

...im not sure what process im supposed to do.

Do just like was suggested: Find any "problem" values, and then declare the domain as all the other values.

Because there are consistent intervals of numbers that cant be in the domain, im not sure how to set up the problem. How do I find what x values of 1 - sin(x) > 0??

Find the intervals were sin(x) < 1. Or have you not taken trigonometry...?

Here's a hint for the limit problem: Divide the numerator by x and the denominator by \(\displaystyle \sqrt{x^{2}}\) It will then be easy to see the limit.

How can I divide the numerator by x when only one of the numerator digits has the x-variable...?

Just do the division, the same as you did back in algebra. Did you try...? The numerator becomes (1 - 2x)/x = 1/x - 2, and the denominator becomes (sqrt[4x[sup:2vuhktpl]2[/sup:2vuhktpl] + x]) / sqrt[x[sup:2vuhktpl]2[/sup:2vuhktpl] = sqrt[(4x[sup:2vuhktpl]2[/sup:2vuhktpl] + x) / x[sup:2vuhktpl]2[/sup:2vuhktpl]) = sqrt[4 + 1/x]. Where did you get stuck?

While you may not like your current instructor, and may even have good reason for your distaste, you do need to at least try. Even if your class notes aren't terribly helpful, you have your book and your knowledge of algebra. Use those, along with the step-by-step instructions you've been given, to at least show how you got started and at what point you bogged down.

Thank you!

Eliz.

 

[Kulith]

Galactus thanks a million, that makes perfect sense to me now, now I can do the rest of the problems like that one.

Are the endpoints of an open interval included within the interval? Now think about the meaning of limits, and max/min values within that context. For instance, if you have the function g(x) = x, is there a "maximum value" on the open interval (1, 2)? Or do you only "get closer and closer" to some (unincluded) maximum value?

The endpoints of an open interval are not included. For g(x) on the interval (1,2) you only "get closer and closer" to some max value.
So, f(x) = 2x + 3 is a linear function with a pos slope.... so I know the abs max is going to be f(2). What about min?? It cant be f(0) because its not included.

Follow the process you've always followed for finding max/min points. Then, as the book's lesson and examples show, also find the values of the function at the interval endpoints (which are included in a closed interval). Compare the various values to find the biggest (this would be the maximum) and the smallest (this would be the minimum).

Hmm i think im starting to remember. The process is using the derivitave of the function to find the "hill's and "valeys" which are max's and min's. Then take those and sub them into f(x) and compare those answers with what you get when you sub the endpoints in for f(x), and you take the largest number as the max, and the lowest number as the min. In this case, the derivative is a horizontal line, so I only use the endpoints to find the max/min? Do I also sub in the endpoint with the open interval?

a. (0, 2]
f(2) = 1
f(0) = -3
absolute extrema (max) = 1
absolute extrema (min) = -3?? or -2.999?

Well, start by reviewing the section in your book, re-reading the explanations and following through the worked examples. It might also help to think about the horizontal asymptotes (for rational functions) that you did back in algebra. "Limits at infinity" are pretty much the same thing. :wink:

yea on my paper it says (HINT: you are finding a horizontal asymptote of the function) but HA to me means finding the derivative and setting the numerator = 0. Im still not sure how its supposed to help me, but I understand how lim x-> infinity is a HA.

Do just like was suggested: Find any "problem" values, and then declare the domain as all the other values. Find the intervals were sin(x) < 1. Or have you not taken trigonometry...?

No I didnt take trig. Took alg1 alg2 geo precalc and calc. Trig was sorta intergrated into precalc. Dont give up on me though.

so the inversse of sin(1) should give me my x values? I get x < 90. Is this supposed to be in terms of pie?

While you may not like your current instructor, and may even have good reason for your distaste, you do need to at least try. Even if your class notes aren't terribly helpful, you have your book and your knowledge of algebra. Use those, along with the step-by-step instructions you've been given, to at least show how you got started and at what point you bogged down.

I am trying man, im trying. You have no idea what im going through with this teacher. Also, posting all this material on a forum takes like 100x longer than it does to write it down and scratch it out etc, ive got like a scrap sheet of paper full of scratcher out attempts.... I got a 660 on my math section on the SAT's so im not a math failure. I am, however, a reading and writing failure lol.

 

[galactus]

For the domain problems. What ever values result in a negative inside the radical are bad eggs.

For instance, the first one:

\(\displaystyle \sqrt{x^{4}-16x^{2}}\)

If you set it to 0 and solve for x you get -4,0,4

Test these values in the expression. Anything less than -4 is good. Anything greater than 4 is good. 0 is also good.

The domain is everything that does not give you a non-real result.

\(\displaystyle (-\infty,-4)\cup(4,\infty)\). 0 is in the domain. It's OK. I am not so sure I wrote the set notation correctly, but you see what I mean.

The interval \(\displaystyle (-4,0)\cap(0,4)\) is what gives you non-real results. So it's no good.

 

[Kulith]

Dude you are amazing at explaining things. I could kiss you.

So for the second one, I set it equal to 0 and then I solve, and I get 90(degrees) or 1.57 (radians) = pie/2

After testing both sides of pie/2 I concluded that all x values worked in the function, and when I graphed it out in my calc it looks like all x values work.

So its (-infinity, infinity)

for the third one, I actually copied it down wrong. I meant, find the domain of f'(x), when f(x) = [sqrt](1 - sin(x))

So, f'(x) = -cosx / (2 * \(\displaystyle \sqrt{1-sinx}\)

Do I still set this equal to 0 to find the domain?? Im lost at this point.

 

[galactus]

I had high hopes for you. But then you went off and spelled 'pi' as 'pie'. :shock: That's a bad sign.

 

[Kulith]

I had high hopes for you. But then you went off and spelled 'pi' as 'pie'. :shock: That's a bad sign.

Ive never even SEEN pi SPELLED OUT, so how was I supposed to know? I dont have the pi simbol on my keyboard btw.

Anyways, pie > pi.

Like I said, I bombed the reading and writing sections on the SAT's

 

[stapel]

No I didnt take trig....

Ouch! That's gonna hurt. You may need to learn trigonometry "on the side", as needed, because it sounds like whatever little was mentioned in your pre-calc course isn't doing the job. :shock:

For a start, you'll want to learn (or review -- thoroughly) the graph of the sine function. For instance, you ask how to find where 1 - sin(x) is negative. But for 1 - sin(x) < 0, you'd have to have 1 < sin(x) or, which is the same thing, sin(x) > 1. What does the graph of the sine function, and the range of values for the sine function, tell you about this inequality? :wink:

Eliz.

 

[Kulith]

Ouch! That's gonna hurt. You may need to learn trigonometry "on the side", as needed, because it sounds like whatever little was mentioned in your pre-calc course isn't doing the job. :shock:

For a start, you'll want to learn (or review -- thoroughly) the graph of the sine function. For instance, you ask how to find where 1 - sin(x) is negative. But for 1 - sin(x) < 0, you'd have to have 1 < sin(x) or, which is the same thing, sin(x) > 1. What does the graph of the sine function, and the range of values for the sine function, tell you about this inequality? :wink:

Eliz.

There is no x value where sinx > 1 ?? So this means all real numbers will work.

can you confirm these answers?

Find the extrema:
a. abs max (2,1)
b. abs max (2,1) abs min (0, -3)
c. none

So for the last problem, the domain of f'(x) = -cosx / (2 * \(\displaystyle \sqrt{1-sinx}\))
First of all, is that the correct derivative of f(x) = \(\displaystyle \sqrt{1-sinx}\))

So Its basically the same thing, just in the denominator. Except now i cant have it = 0 either. So at PI/2 i get the denom = 0.

So the domain of f ' (x) = (-infinity, PI / 2) U (PI / 2, infinity)