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#1 Re: Exercises » Domain and Range » 2018-11-17 20:49:45

The domain is the set of possible 'x' values and the range is the set of possible 'y' values.

N15mrei.gif

You are correct that the curve tends to zero as x tends to infinity but that doesn't tell us the range.  You cannot have the value x=3; it is excluded from the domain, but you can get as close to 3 as you like and as you do the function has larger and larger values.  There is no value of 'y' that cannot be attained so the range is ( -∞  , ∞ )  Note y = zero is attained at x = -2

Bob

#2 Re: Exercises » Domain and Range » 2018-11-14 21:36:33

hi

There's a lot of brackets there so here's how I analysed the function:

(SQRT(x+2))/((x^2)-9)

(SQRT(x+2))      /      ((x^2)-9)

Drop unnecessary brackets

SQRT(x+2)        /       (x^2-9)

Question 1.  Is this the correct function ?

Question 2.  Have you looked at the graph linked in post 2 ?

Please answer these questions.  If I don't get both answers I shall not be posting again on this thread.

Bob

#3 Re: Exercises » Domain and Range » 2018-11-14 03:43:51

x                           F(x)
     0                     -0.15713484
  100                0.00101086
   -1                    -0.125
    2                    -0.4
2.99999            -37267.82447
2.999999999    -372677965.4
3.00000001     37267799.89
3.000000001     372677965.5

Using Excel to calculate =SQRT(x+2)/(x^2-9)

As x approaches 3 from below the values of F are getting increasingly big in the negative direction.
As x approaches 3 from above the values of F are getting increasingly big and positive.

Have you looked at the graph?

Are we using the same function ?

Bob

#4 Re: Exercises » Domain and Range » 2018-11-12 20:07:10

hi Zeeshan 01

I'm not sure if there is another way.  Have a look at this page:

https://www.mathsisfun.com/sets/domain- … omain.html

Somehow you have to find all the possible values that F(x) can have.

You cannot list them in a set as this is (except at x=3) a continuous function.  So you have to write the answer in this way:

[lowest values, highest value] although sometimes a square bracket may be replaced with a round bracket to indicate that an endpoint is not included.

I think that considering the graph helps a lot with this.  Did you follow the link?  It shows that the function is continuous in two sections separated by x=3.  It also shows that the function tends to both + and - infinity.  Thus we know that the range is all real values of x.

Bob

#5 Re: Exercises » Domain and Range » 2018-11-11 20:34:00

Domain f(x)= [-2,+infinity)

Yes, but you must also add x  ≠ 3 as there is no value for F(x) at this point. infinity is not regarded as a number.

The range is all the values that F(x) can take.  (x-3)(x+3) determines this because when the denominator approaches 3, the denominator tends to zero so the function tends  to either + or - infinity.  This means that all values occur for F(x) so the range is (-∞ , +∞) If you look at the graph, choose any value for F and draw a horizontal line to try to cut the curve.  It is always possible.

Bob

#6 Re: Exercises » Domain and Range » 2018-11-10 20:52:46

hi Zeeshan 01

Haven't heard from you for a while.  How are you doing?

To answer your question, I made a sketch of the graph.

Firstly, if x < -2 then the square root cannot be computed so that sets a lower limit and determines the domain.

At x = -2 , F(x) = 0

x^2 - 9 = (x-3)(x+3) so there will be vertical asymptotes at -3 and +3, meaning that x tends to either + or minus infinity. The negative one is outside the domain so we needn't consider it.   Consider what sign F(x) has as x approaches 3 from above and below.  When x > 3 and approaching 3 both the numerator and denominator are + so the curve will tend to + infinity.  When x < 3 and approaching 3 the denominator switches to negative so that part of the curve tends to - infinity.  That's enough to decide the range.

If you want to view the graph go to https://www.mathsisfun.com/data/functio … 2)/(x^2-9)

Hope that helps,

Bob

#7 Re: Help Me ! » Law of sines and Trig » 2018-11-05 20:11:20

hi careless25

Angles PQS and SQR are shown with two arcs.  I think this can be interpreted as meaning that they are equal.

So in triangle PQS that angle can be found and then in triangle SQR you know two (and therefore all three) angles and a side.  So x can be found.

Bob

#8 Re: Help Me ! » How do you get v dv/dt = d/dt (v^2/2) » 2018-11-01 23:48:55

hi kpstatic

Welcome to the forum.

In mechanics, to switch between acceleration, a, velocity, v and distance, s it is necessary to differentiate and integrate as

I seem to remember this bit of algebraic manipulation arises when dealing with simple harmonic motion, because a is defined as a function of s rather than t.  So a way is needed to express a in terms that can be integrated.

If you post back more precisely where you've got to in your course and why you came searching here, I'll try to provide more help.

Bob

ps.  s is used for distance as d is likely to be used for differentiating. x may also be used.

#9 Re: Help Me ! » Volume of irregular polyhedrons » 2018-10-20 19:51:35

Hi Mathegocart,

Thanks for the help.

I'd forgotten I'm on someone else's wi fi at the moment. It's their content blocker.  Seems to be a common problem with Vodafone. So I'll wait and see if the OP wants my diagram before tinkering with the settings.

Bob

#10 Re: Help Me ! » Volume of irregular polyhedrons » 2018-10-19 21:34:57

hi thehay95

Welcome to the forum.

I've had a go at sketching this solid.  Trouble is that just having the letters PQRSTUV doesn't exactly tell me which points to join to which.  It looks like there's a prism PQRSTV, which has a triangular cross section QRV, and sitting on this is a rectangular based pyramid TVRSU, where the base is TVRS and the vertex U.

Volume of the prism should be fairly straight forward.  Calculate the area of the triangle QRV and multiply by the length, PQ.

The pyramid is more tricky.  You'll need the area of the base, TVRS.  This is a rectangle.  You need Pythag. to calculate VR.  Then the perpendicular height from TVRS to U.  If you draw the triangle UZY, you'll see that UZ is not perpendicular to ZY so that length won't do.  But you can calculate UY and the angle UYZ and so treating UY as a hypotenuse you can get the required height by opposite (height) = hyp x sin(UYZ).  Once you have this the volume of the pyramid is  one third of the area of base times the perpendicular height.

Then just add together the two volumes.  smile

Hope that helps,

image upload problem.  I have a diagram but I cannot log in to imgur to upload it.  Some problem with a Vodafone security certificate ??? I'm on BT broadband.  Anyone got any suggestions.  It happens with both IE11 and Chrome. My account still exists as old images are still showing on the forum

Bob

#11 Re: Introductions » Introduction » 2018-10-08 19:23:24

hi iamaditya

There's a thread on the forum that includes the discussion and the link to the download.  It was a while back and I've forgotten the details; sorry.

Bob

#12 Re: Introductions » Introduction » 2018-10-07 19:53:06

hi iamaditya

No Spam. Spam includes messages that have no relevance to the topic, that are annoying, repetitious or promotional in nature.

I remove adverts because they are (1) sometimes pornographic; (2) they 'clog' up the site with irrelevant rubbish; and (3) it's not fair to the paying sponsors to allow free advertising. If they stop supporting us the forum's very existence  is at risk.

Mostly, these posters join the forum; create one or more posts which contain an advert, either in the post or the signature, then we never hear from them again.

Just once, we had a genuine mathematician who posted a link to the sale of his maths book.  Bobbym had a word with him, and he removed the advert and replaced it with a link to a free download of the book.  That seemed like a good compromise.

Bob

#13 Re: Help Me ! » Algebra: Polynomial Question » 2018-10-01 19:34:49

hi math9maniac

The 'simplification' was not valid.  √ (A-B) is not generally equal to √ A - √ B

I squared both sides of the equation to get

I'm still not seeing any real solutions.  See https://www.mathsisfun.com/data/functio … ^8 - x + 2

Bob

#14 Re: Help Me ! » 3d plane in a cube » 2018-10-01 01:28:18

hi n0tjyxgsgs

Welcome to the forum.

This problem came up ages ago so I've mostly forgotten it.  Some other posters provided quicker solutions than mine; that's fine; I never said mine was the only one nor that it was quick.  The OP was happy to be taught some new maths along the way which I was happy to do.

PQRC is a sloping plane.  So QR isn't a continuation of PQ.  Not sure what you mean here.

Bob

#16 Re: Help Me ! » Question » 2018-09-29 21:04:05

Hi Sterling94,

Welcome to the forum.

When two dice are thrown there are 36 possible outcomes but we can limit ourselves to just (5,x) 6 possibilities plus (x,5) another 6 possibilities less the one we've counted twice (5,5).  So 11 in total.  Of these two give a total of 8 so 2/11

Bob

#17 Re: Computer Math » Software Solution 2019 Latest Updates » 2018-09-27 19:19:50

hi joseph

Welcome to the forum.

Thanks for the supportive comments.  MIF is funded by carefully selected advertisers.  Please don't use the forum for free advertising.

Bob

#18 Re: Help Me ! » Trigonometric Confusion? » 2018-09-27 19:16:47

It's very odd.  You'll have to raise this with your teacher.

B

#19 Re: Dark Discussions at Cafe Infinity » whats 1+1? » 2018-09-26 19:24:24

Arhh! Good old binary answer.  Or maybe 1 + 1 = 1 as in one pile of sand added  to  one pile of sand = one pile of sand.

Bob

#20 Re: Help Me ! » Trigonometric Confusion? » 2018-09-26 19:20:45

hi Mathegocart

Typos can creep into questions.  In the UK the national exams are, nowadays, subject to a load of checking but question errors still get in sometimes.  [eg. https://www.independent.co.uk/news/educ … 27141.html ]

Or maybe it was intended to test if you are aware that sometimes a calculation cannot be completed.  If I was going to do this I would word my questions (have to be several) "Work out all the missing side lengths and angles in these triangles.  If you think a triangle cannot exist explain why."  Something like that would be reasonable if it was included like that as part of the assessment.  If it was a single question on the topic, and you have copied it correctly**, then it looks like a serious error.  What was the test?

Bob

** If the numbers are swapped around then it is possible for example.

#21 Re: Help Me ! » Trigonometric Confusion? » 2018-09-25 19:38:21

hi Mathegocart

No error.  No triangle can be made with those measurements.  Try it yourself.  Draw AB = 27 (may need to scale down) and angle B = 38.  You don't know how long to make BC so just continue the line for some distance.  Set a compass to a radius of AC = 15 and try to intersect BC.  This fails.  The shortest distance from A to BC is about 18.

Bob

#22 Re: Guestbook » 0.999... = 1 :: alternative version » 2018-09-07 19:42:21

hi Xepemu

Welcome to the forum.

That's a neat method which seems to overcome the objections some folk have about infinite digits.

How about joining the membership?

Bob

#23 Re: Exercises » Percentage » 2018-09-07 19:39:38

hi Sresta702

Welcome to the forum.

Which ans?  I like to encourage posters to work things out for themselves.

Bob

#24 Re: Introductions » Hi there! » 2018-08-25 19:45:40

hi Pablo,

Welcome to the forum.

Bob

#25 Re: Help Me ! » Does 1 = .999...? » 2018-08-25 19:42:08

The difficulty arises because it is a property of the real number system that between any two real numbers there is another real number.  For all other reals this works ok but it fails whenever the decimal representation 'ends' with an infinite string of 9s.  I have encountered two 'solutions' to the problem and both allow a consistent set of axioms.

(1) Allow that for example 0.99999999.... is the same as 1   I can understand why some people dislike this but, if you remember that the reals exist irrespective of the decimal representations, then why not?  Lots of mathematical topics work perfectly using this solution.  For example, you can convert an infinite, and recurring, decimal representation into a fraction.

(2) You can forbid the existence of such numbers from the real number system.  That also works and is the basis of the approach used by Georgi E Shilov in his book Elementary Real and Complex Analysis.

It might help if you stop thinking that numbers have a concrete existence and accept that they are just convenient abstract ideas that help us to do certain types of mathematics. (You can hold 3 apples or even a wooden shaped '3' object but you cannot hold a 3.)  So we can make numbers do what we wish according to the model we are making.  Not everything obeys the rules for real numbers.  For example add one pile of sand to another pile of sand and you have one pile of sand. You cannot have half a stick of chalk.

In summary, decide what mathematical model you are building; invent a consistent set of axioms for your model; and then use it to discover new things.

Bob

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