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LaTeX is basically the way that all math in academia is typed, its ubiquitous everywhere from math to physics to engineering to computer science. Check out Overleaf for some examples and a very convenient online latex editor.
Start by factoring out an x^3 in the first one, and x^4 in the second one. You'll have like terms after that, because an x can come out of each radical. Then use the sum rule for limits, factor out the x, and then use the product rule for limits on each. At that point, you'll have 2 limits of polynomials, and 2 with nth roots. Then use the root rule for limits on the roots.
The nice thing about factoring both of those radicals is that it puts powers of x in the denominators of the other terms, and all of those go to zero as x goes to infinity. This is what we want.
Here is the graph:
https://www.desmos.com/calculator/cjiecnczde
Appears that the limit is a very slow infinity.
This is an application of the Fundamental Theorem of Calculus.
As you probably know, if we want to find maxima/minima we should look where f'(x)=0 or DNE.
The derivative of an Integral is simply the Integrand function itself so we have that: f'(x)=e^(xcos(x ) (cos(x)-x(sin(x))
So when is this zero? A few places actually, but the solution is only available numerically since the equation is a mix of polynomial and transcendentals.
Checking our trusty Desmos to find these things for us we see that at around 6.437 we do indeed have a maximum for f(x) and that this is where f'(x)=0. Note that in addition to being a maximum here, we notice the derivative function satisfies an Interval test of some sort to confirm that we do indeed have a max.
I'd say polynomials are an easy bet. Start with them in standard form, take some derivatives and make some systems of equations for the coefficients.
Also, don't be afraid to fool around in something like Desmos. This took me like 5 mins:
https://www.desmos.com/calculator/wr2r3qz883
I mean, the curves are terrible as far as anything physically possible, but it is do-able. But, if we go with some horizontal stretching, blah blah, it does get better. I guess trig functions are great, too.
Dunno - randomly firing off ideas. Useful?
Just recall that for limits, we have a product property that tells us
lim(x->a) f(x)g(x) = lim(x->a)f(x) * lim(x->a)g(x)
Using this, we can let f(x)=ln(x^6 - 7)/ln(x) and g(x) = 1/cos(1/x).
The latter function has limit 1 as x->infinity, as 1/cos(1/inf) = 1/cos(0) = 1/1 = 1.
We can solve the limit as x->infinity of the first function using L'Hospital's rule once.
The end result is 6.
The graph of this function verifies this result: https://www.desmos.com/calculator/fi0qa9zhg2
Written solution: https://i.imgur.com/B8otXHO.jpg
Edit: Aplogies for putting x^6 - 1 rather than x^6 - 7. My typo. That constant following x^6 has no effect on the final answer, regardless of what it is.
The Function approaches 1/2 as x approaches infinity. The red function is your original function. As you scroll further and further to the right, it gets closer and closer to the blue line fixed at y=0.5, but never reaches it.
EDIT: Formatting
I know in general the chain rule and product rule can solve many and most derivatives.
Your thinking is pretty good!
I've included a graph for you below here to illustrate the problem, and I think you'll notice that the Integral is indeed only defined on (-3,3). That said, given what you mentioned about where you are in the course, I kind of doubt you could give a great reason why intervals such as (-inf,-3), and (3,inf) shouldn't be included either.... something I attribute to a somewhat unfair question.
That said, what's nice about this problem is that a closed form anti-derivative does indeed exist, and I've included that in the graph too for your comparison.
Note that one easy answer to when an integral is equal to zero is simply to make the lower and upper bounds of the integrand equal to one another: you take a slice of zero width.
So, to answer b) you can simply say x=1.
Sure.
PART 1:
(sqrt(3)-i)^5 = ∑C(5,k)(sqrt(3))^(5-k)(-i)^(k) = -16i-16sqrt(3) This means the -b/4=4
g(2)=3, g'(2)=2 → y=2x-1 is the tangent line equation. This means c+3=2
L'Hopital applied twice does it, or Taylor Series applied once(my preference) implies the limit must be zero. Confirmation: https://www.desmos.com/calculator/cqyjgtd9if
PART 2:
Umm... quadratic equation.
Just like question 3 above.
It's Analytic so you know the limit will automagically be 1, but barring that observation, do logarithmic diff'n.
Substitution: u=x^2
Tedious but straight forward
Pick a suitable constraint function and go.
- i haven't paid much notice to hyperbolic things in a while so i didn't try the others, but the ones above... didn't take much time. Really the annoying part is dealing with the frankly ill suited response format. So weird.
EDIT: Sorry about the numbering, apparently reddit applies autonumbering despite actually picking different numbers for each question. If anyone is unsure or would like to see a solution, just let me know!
Edit:
Here are the answer sheets! https://drive.google.com/folderview?id=0B-P-7f3GdG51U3FWQVA1d0ItUUE&usp=sharing
There are probably three questions that I might've done incorrectly. If you spot any wrong answers, please let me know and I"ll be sure to correct them right away! Thanks for checking it out!
I’ll add my 2 cents: this may help, get the book “humongous book of calculus problems by Michael Kelly” (see link below).
You may brush up on the fundamental theorem/topics of your choice to prepare you for Calc 3 multivariable calculus.
I would start by doing the review problems at the end of each chapter that you went over in class. Do all the problems that seem to fit the assigned problems in homework.
Two resources I'd recommend. Chegg app This wolfram app
http://i.imgur.com/bpH1ArY.jpg
I'm not sure where the = x^2 comes into play, but watch these videos to follow my steps.
http://www.khanacademy.org/math/algebra/polynomials/dividing_polynomials/v/algebraic-long-division
Here is a good video from a great teacher. FYI The antiderivative of (e^u)*du is just e^u. That's in the video and you probably haven't covered it yet.
http://www.khanacademy.org/math/calculus/integral-calculus/u_substitution/v/u-substitution
So I retook calc 2 last year, in a 6 week summer course, and it was my first math class in 10 years. I needed to reteach myself calc 1 in that time too. One tiny pocket sized book saved me.
http:/patrickjmt.com , first and foremost. https://www.coursera.org/courses?query=calculus (free online courses in calculus I,II,III that you can take at your own speed)
Also remember a lot of calculus concepts are easy to recall, however students tend to make small mistakes with algebra. You should be sure brush up on your algebra as well. PatrickJMT has everything thing you need to know. He was a god-send for me.
Best of luck to you!
I originally posted this in math stack exchange but got no answer if it is correct. I am now pretty sure this is correct as by looking at the desmos graph https://www.desmos.com/calculator/v0uulbfbsa , you can see that both the graphs match in the range of e^(−e)<x≤e^(1/e). This is the same range at which the integral of x^x^x... is defined. Could someone help me verify if this is correct?
​
Well, the series converges by the limit test, and you can look at the "graph" to verify:
https://www.desmos.com/calculator/8caiz5d1dm
The graph shows why the series has to start around n=3.
The ratio test also gives zero.
Do you have to do the root test, or can you do any test you want? These two here appear to be simpler? Does this help?
Everything you did is great. Please confirm that the second derivative changes sign, not just that it is equal to 0 though. Here is a graph of your original function with the inflection point highlighted, and you can see that the graph switches from an "n" shape (concave down) to a "u' shape (concave up) at x=2/3. https://www.desmos.com/calculator/boafwe8tm9
I assume that the log has to be perpendicular (ish?) to both shores.
Maximize the distance formula (https://en.wikipedia.org/wiki/Distance#Geometry) for this one. The two pairs of coords come from:
y1=(x1+1)^2 and y2=5(x2)^2 +2.
Specifically:
(x1, (x1+1)^2 ) and (x2, 5(x2)^2 +2)
Here is a graph of the situation:
Graph the polar pattern, then convert the rectangular equations (y=4, y=0) to polar.
Limits of integration are basically from theta = 0 to pi, and r = 0 to 8 + 8 sin θ.
The integrand should be the outermost curve minus the innermost curve. Seems that the outermost is the cardioid and the innermost is the line for the front of the stage. This will be the parts of the system bounded by the front edge of the stage and the cardioid pattern. This is analogous to upper curve minus lower curve in rectangular coords, but we have r to consider, which is radial, so outer minus inner is the same kind of thing.
Graph is:
https://www.desmos.com/calculator/ojb2seby7j
Does this help?
+/u/dogetipbot 100 doge verify There you go man :) If you don't already know what it is, check this out.. After that, head here to download a wallet. Also check out /r/dogetipbot to learn how to work tips on reddit.
Go to Z Library. You can find plenty of calculus text books and their solutions manuals in PDF format using the search function. The most popular texts are probably either Stewart or Larson.
You're most of the way there. You've got your derivative right, and the 0 of the derivative on that interval is pi. The last thing you do is plug that value into the initial function, which gives you your critical value, which is 2. This is part of the answer to (b). There's also a minimum at x=0 and x=2pi, which is -2. This graph may help: https://www.desmos.com/calculator
It's just the definition of the derivative where x=1. I suppose the tricky part is in quickly realizing that √1=1 so that the expression as written really is the derivative of the square root function for x=1.
So the answer is simply f(x)=√x and a=1. See this for what might be some help: https://www.desmos.com/calculator/cess5sbq9q
A double integral's bounds describe area (again analogous to the single integral, whose bounds describe a length). As /u/odog_ says, you (or your professor!) need to be careful when switching the order of integration so that you're describing the same area.^(*)
The bounds should indeed be from y to 1 to get the same area:
https://www.desmos.com/calculator/8ni61alror
I believe you've made a calculation error. Swapping the bounds of integration always negates the original value, even with multiple integrals. Here's the integral evaluated from y to 1:
^(* Technically, this should be signed area, but I doubt you'll face a problem where the area is negative, just as you're usually not integrating from 2 to 1 in single-variable calculus.)
Regarding the function: apparently you can do subscripts on this subreddit.
f(x) = Σa*i*x^(i)
turns into
f(x) = Σa*i*x^(i)
Hopefully this form is a bit clearer.
This is how I would represent polynomial function of degree n in any context. i takes the values from 0 to n. A verbal statement of it might be as follows: "f(x) is the sum of various monomials of x, i.e., x raised to some constant power and multiplied by some constant coefficient." Or, unrolled (summation -> sum), the function would be:
f(x) = a*0x^(0) + a1x^(1) + ... + an*x^(n)
which may be the more familiar form of a polynomial.
~~As for coding: Any popular programming language ought to be capable of automatically fitting a polynomial curve to a set of points, and have some tricks to do it more efficiently. As I said, it'd be a one-liner to do it in Python.~~ Or you can use Desmos's curve-fitting abilities, as demonstrated with the bottom part of the eye here.
Sorry, west coaster here, knocked out for awhile. (APs came out. D:)
I have attempted to craft a set of points that mimics the top of the eye here.
Now, we can fit this set of points to a polynomial function. Our function can be represented as
f(x) = Σ^(i)ax^(i)
where ^(i)a is the ith coefficient, corresponding to x^(i).
The function needs to contain the points, so let's plug a point in and see what happens. Here's (1, 1.1):
1.1 = Σ^(i)a(1)^i
Σ^(i)a = 1.1
^(0)a + ^(1)a + ... ^(n)a = 1.1
Notice that this is a linear equation.
Now, what is n, the degree of the polynomial, assuming ^(n)a ≠ 0? Well, again, we need a polynomial of N - 1 (oops, typo) degree to represent almost any set of N points. You can verify this for N = 2 by remembering that two points make a line. In this case, we have 9 points, so the degree is 8.
A degree of 8 means 9 coefficients (remember the constant). We thus need 9 equations to come up with a unique solution. And we have those - as just shown, we can denote the points as (x, f(x)) to come up with an equation. Another example - (2, 1.5):
f(2) = Σ^(i)a(2)^(i) = 1.5
Now, solving a 9-rank system isn't very fun, so here's some Python code to do it:
I pasted the list of coefficients into Desmos and made the function from it.
And there you have it.
(Note that this is a one-liner in Python if you use pre-made code.)
e: fixed link
This is incorrect. You seem to be misunderstanding what an indeterminate form is. The expression 0^(∞) is not an indeterminate form, since it is not indeterminate (it would always correspond to a limit of zero).
All indeterminate forms are notated using technically incorrect notation, but they are expressions that represent limits of functions that respectively have limits of 0, 1, or ∞, where the limit could plausibly take various values, but we do not yet know which. For example, "0/0" as an indeterminate form represents L=lim_(x→c)f(x)/g(x), where lim_(x→c)f(x)=lim_(x→c)g(x)=0. It doesn't mean that we're actually computing 0/0.
Indeterminate forms represent a subset of such combinations of functions. For example, in the previous example of 0/0, with c=0, we could have f(x)=g(x)=x, which would give L=1 or we could have f(x)=x^(2) and g(x)=x, which would give L=0. When we say that the limit "has the indeterminate form 0/0", we mean that we are not sure of its value.
Not all such combinations of functions are indeterminate forms. The expression 1/0, representing lim_(x→c)1/g(x) is not an indeterminate form, since it always takes the value ±∞ regardless of the function. This is also true for 0^(∞), which represents lim_(x→c)f(x)^(g[x]), where lim_(x→c)f(x)=0 and lim_(x→c)g(x)=+∞. Regardless of our choice of f and g, this expression will always be zero.
See (Brilliant, "Indeterminate forms", Forms that are not indeterminate) and (Wikipedia, "Indeterminate form", Expressions that are not indeterminate forms).
From a fellow ADHD comp sci student who takes SO LONG to complete math/physics problems, different colours are a lifesaver.
I've recently started using my Windows Surface primarily for math, which makes this somewhat easier but, before that, these were my go-to for math & physics notes and homework.
Word problems mess me up in particular and, as another user mentioned, giving parts of the problem different colours helps a lot.
Good luck with it!
>email the image to my tutor.......the tutor could take a look at the image (i.e.: take a look at my work that way).
That is a process that slows down the tutoring experience altogether.
If you had a document camera of your own, you would be able to share the live feed to your tutor. Zoom would just detect it as another web cam that you could switch to during the meeting.
This looks like a suitable option. Yes, it’s a bit expensive, but it will be totally worth it if you work with the tutor regularly throughout the term or year.
Just set up a well-lit work space for you to do your work with the document camera set up to capture your work.
In my course we looked at finite difference methods up to the 2nd derivative. Upwards of that was never necessary (Mostly because it was focused on physics - we use it for the Schrödinger equation but that's it.)
However, the book we used has some information on higher order derivatives as well. It was this one.
Word out there is that you don't necessarily have to pay for it if you know what I mean.
OKAY So I fiddled around again with the previous solution I got, and I realised there is nothing that technically restricts us to assuming cos(u)=1 specifically. We could always just assume cos(u) is approximately a instead, and then move the area cover technique so it covers the entire plane. Well, I managed to pull it off in desmos and it now seems at least functional. I think if you could formally take the limit of this process, you would get your actual function. Here is the desmos page.
https://www.desmos.com/calculator/1k83rteor3
I only used 6 coverings in this, and you can enable the actual areas if you go to the "areas" folder. It gets messy though. You will also notice that the current "function" we have is quite discontinuous. Anyway, hopefully some of these musings help in some way, although I doubt it.
Yeah, you can use Desmos to graph the function inside the integral and the bounds.
-----
For example, if you had...
∫ [x + sin(x)] dx
...with bounds from 0 to π, then you would graph the function in the integral as y = x + sin(x) and you would graph the bounds as x = 0 and x = π. This would show you the general shape of the region you are finding the area of.
For the first part, solve each of those equations for y, and those will be your f(x) and g(x). Find where those graphs intersect, and those are your a, b, and c values. This is the hard way to do this problem.
​
The second part is the easier way. Solve each of those equations so you have a function of y instead of functions of x. You should see that, with respect to y, one function is always on the top and the other function is always on the bottom, making the difference of those functions easy to compute and assign to h(y). You can use your results from the first part - the a, b, and c values - to develop what your alpha and beta values should be.
​
Here is a graph of the situation:
https://www.desmos.com/calculator/1n1dicr1is
​
HTH
Nope, that's the only point they give. That was part of my confusion also. The due date has now passed, so I accept my fate haha. Here's a desmos you can look at the I used to check my answer originally to confirm that m=-2 https://www.desmos.com/calculator/3lkq5hae2h
This is a Fundamental Theorem of Calculus (FTC) question. The key is the phrasing of that integral on the left-hand side.
Take the derivative of both sides. For the left side, that means dropping the integral and replacing theta with alpha. For the right, it's derivative with respect to alpha.
Here is the graph after derivatives: https://www.desmos.com/calculator/3ns4pxqoxy
Clearly, the red curve is above the blue one for (0, pi/2).
​
Take a derivative, then set it equal to zero and solve for x. Will be one value. You just have to check what's happening with the derivative on either side of that value.
Here is a graph of f and f':
https://www.desmos.com/calculator/ocvkqe1b7o
HTH.
Have a look at this graph, which identifies the tangent to the skinny parabola at its minimum. Obvious, of course, but what it does is show where the "bend" really is (I stand by (0,2) needing to be fixed) as well as show what domain we have to consider for the outside parabola, namely, where the outside parabola intersects y=2. Minimize the distance function over that domain of values, and I think that's where the answer is.
https://www.desmos.com/calculator/jj2fs9bnu0
The graph shows that the "bend" is only between the min of the outside parabola and y=2. Else, the shores are "opposite" one another, yeah?
Looking at the graph: https://www.desmos.com/calculator/gjvsjlojhy, -1 is not a root. The root appears to be rational and between 1/2 and 1. Since this is a quintic polynomial whose graph crosses the x-axis only once, we have one real root and two conjugate pairs of imaginary roots for this function.
Have to do b) by THREE integrals.
To the left of the y-axis, radius is 7-x^2. Interval is [-sqrt(7), 0].
To the right of the y-axis, radius is 7-(4x-x^2 ). Interval is [0, 2].
On the interval [2, sqrt(7)], the radius is again 7-x^2.
Add 'em up.
Watch those parentheses! Always.
The graph is useful: https://www.desmos.com/calculator/cvscj9cxo7
Pretty sure this is correct. Have a go?
https://www.coursera.org/course/calcsing
Single variable online free calc course. Starts off talking about Series and according to my roommate who already watched all the lectures, Taylor series is brought up all the time. The course doesn't technically start again for 2 months, but all that means is that you cannot do the homework, but you can see all of his lectures
Techniques of integration and sequences and series are the main staples of calc 2, I also took and taught this class with Stewart.
Here’s what I recommend, buy Schaum’s Outline of Advanced Calculus for $20 on Amazon for me it shipped in a day here in California. Study Chapter 5-Integrals first, as it is likely the first set of material in your Calc 2 class, then skip to Chapter 12-Improper Integrals next, then skip back to Chapter 2 -Sequences, lastly skip to Chapter 12-Infinite Series and that’s Calc 2 in a tiny nutshell, btw the Chapters were based on the 3rd edition. Text I had you buy for cheap will also help you with calc 3&4 if your calc goes to 4 that is. Don’t just read through the lecture potion of the book, do work lots of exercises in integrals, sequences and infinite series.
As a special treat, you may wish to study the gamma and beta functions in chapter 15 and you study Chapter 12-Improper Integrals. This material is useful and interesting, challenging and should be considered as extra credit or the cherry on top. Up to you—the gamma function is the extension of the factorial function to real numbers less the negative integers in this book. I mean the factorial that works on integers
4! = 4x3x2x1 = 24.
Good luck and PS don’t let the Advanced in the title scare you it’s just regular calculus with some extra topics included like Gamma and Beta functions (the gamma function made a cameo in 9th Ed. Stewart Calculus: Early Transcendentals, but not at the level of Schaum’s.
If you get stuck just post on here or msg me and I will help I have a copy of the text even. That’ll work if I’m paying attention. If you have the Line chat app or discord chat app I can give more reliable means of getting ahold of me.
looked up on worldcat, don't have time to go through them so here's the link
Does every number of the form x^a , where x < 0, have to be represented by a + bi, even if b = 0? So there's always an i there, which makes things undefined on a non-imaginary plane, therefore there's no graph for x < 0.
The graph of <em>y</em> = (-2)^<em>x</em> has only one point: (0, 1). Here's what I think Desmos is doing: Desmos (or any graphing calculator) sees a negative base, thinks that for any x ≠ 0, y must be in the form a + bi, says "shoot, there's an i, so I can't graph it!" and proceeds to graph only (0, 1).
Well, you can tell that x=0 is a critical point. x^2 will dominate sin(x) so there's only going to be one more chance for a critical point, which you can't find analytically. But it has to be when 3x^2 is less than 1, so x^2 < 1/sqrt(3). So will a bound for the other critical point be good enough? cf https://www.desmos.com/calculator/i0biicdehu
Thats exactly what im trying to do thank you for the video. However this video dealt with a very simple function "sin(x)" my function in which im approximating is rather complex. https://www.desmos.com/calculator/d0pshy1nhx
honestly, I believe a simple regression would do, however my assignment requires me to use calculus to approximate an equation. is there a way to use the Taylor Series on a scatter plot as a regression..... without having a pre-existing function like used in the 3Blue1Brown video....
The basic algebra is essential. You'll likely not be able to use a calculator with your professor, so be sure that you can evaluate simpler formulas "off the top of your head." Some important bullet points - - Factoring - Understanding roots / exponents [ex, sqrt(3) is 3^1/2]. Logarithms inclusive (sadly) - Completing the square. Not used often, but needed in some cases. - Sum / difference of squares/cubes - Transformations - Understanding the "library" of functions (constant, linear, quadratic, cubic, sometimes quartic, sqrt, abs, and exponential.) - Formula manipulation, but that's part of the algebra... - Use radians if you don't want to off yourself mid-semester. - Trigonometry | Including identities
Desmos will be your friend.
OK, the graph of this one is insane. Load this up and zoom in. On the interval [-0.2, 0.2], we have problems.
https://www.desmos.com/calculator/yf5qiazn25
Something is telling me that the derivative of this thing is very problematic. The derivative itself (if I've done it right, using quotient rule) has that oscillating - limit doesn't exist near zero - type of thing going on. The function, f, also has weirdness going on on the interval I mention.
I've tried a bunch of tricks to resolve, but you're right -the h keeps on hanging around. Is this first semester Calculus?
Tough one.
I would first start with finding a program that can draw dots on an X/Y plane and do equations. I recommend JSFiddle.
I get you're into Naruto and you want to show off, but if you're not good with coding or drawing and you can't outline what curve goes with what equation, then your assignment should be very simple.
Are you familiar with any kind of programming?
Yes it is. Most trigonometric equations will have an infinite amount of solutions on all of R. Hence you want to find all solutions in the given range. Your calculator will automatically assume your solutions are in the range of [0,2pi), so you will have to find the other solutions. The rules for finding more solutions is as I said sec(x) = sec(x +2pi). The same counts for sine and cosine. For tan the rules are tan(x) = tan (x + pi). If you need help with remembering, you can use this calculator: https://play.google.com/store/apps/details?id=cz.hipercalc which I greatly recommend. It will show you what radians you'll have to rotate your solutions with. Not sure you can use it on an exam thought.
Honestly I'm going to be a jerk and say yes, I'm asking for the work. I'm having a hard time wrapping my head around this all. I'm trying to use this link http://www.symbolab.com/solver/definite-integral-calculator/%5Cint_%7B0%7D%5E%7B4%7D%20xe%5E%7B(-x)%5E2%7Ddx/?origin=button
Desmos folks helped me punch up the graph: https://www.desmos.com/calculator/srcmzxqha5
I was also stupidly forgetting polar area, so the integral is $.5 \int f(\theta)^2 d\theta$ because the base unit is triangles from origin, r long and r dtheta wide.
That makes it analytic by substitution, http://www.wolframalpha.com/input/?i=integral+(sec(x)tan(x)%2F(1%2Btan(x)%5E3))%5E2+dx; hopefully you can see the u in the result!
trig sub is not necessary. u-sub then arctan does it:
∫dx/√(e^(2x) - 1) = arctan(√(e^(2x) - 1))+C
https://www.desmos.com/calculator/9xjvwjcdrw
edit: following up to your last comment, they're really the same function.
I would suggest using a graphing a utility (like desmos) to help visualize what is required to make f(x) continuous.
I would also search youtube for some videos about continuity, if your nephew doesn't understand the topic. Just giving you the answer won't really be helpful and goes against the mission of this sub.
It isn't necessary, particularly in this example.
Pick an n, subdivide your region into that many slices, find the area of each slice, add them up.
I've been particularly lazy here and chosen n=1: https://www.desmos.com/calculator/z5ftftsxjo
The blue region(my Riemann "sum") area is just sqrt(3/2)*1=sqrt(3/2). This is about 1.22 The correct area for the curve is the red region, for which the area is also 1.22, correct to 2 decimals places.
So they match very well, despite the highly lazy approach.
You're right. The answer is either -4 or DNE.
The DNE answer is correct in the sense that we don't see the limit as x approaches -2 from the left, so the limit is one sided and therefore the overall limit DNE. However, this is really down to the squeezed windows size used in the question. Does the question intend to do this? Hard to say.
-4 is the correct answer if we assume the function is continuous in some small neighbourhood around x=-2 since the limit of a product of 2 continuous functions is the product of the individual limits, and (-2)^(2)*(-1)=-4
Incidentally i created what i believe is the actual equation of the function in question. Check it yourself to verify the limit (if it exists at all, see DNE thing above), really is -4:
For sure there is, especially when you are familiar with a couple basic MacLaurin Series with which to compare.
You know that 1/(x+1) = ∑(-1)^(n)x^(n)
The idea here is to exploit this knowledge and avoid having to do all the derivative calculations. In other words, make your question look like the form of one you already know. Here's how:
Let z=x-2 → x=z+2 Then:
1/(x+1) = 1/(z+2+1) = 1/(3+z) = 1/{3(1+z/3)} = (1/3)*[1/(1+z/3)]
This last form is exactly what we want.
Since: 1/(x+1) = ∑(-1)^(n)x^(n) Then: (1/3)*[1/(1+z/3)] =(1/3)∑(-1)^(n)(z/3)^(n) =(1/3)∑(-1)^(n)((x-2)/3)^(n)
And this is exactly what you want. No derivatives required. The math syntax here is pretty horrible. This link should clear up any of the notation and provide a quick graphical check of things:
Do you know how to find points on a function? Draw your axis and plot some points, connect the points with a line. You'll usually be concerned about your x and y intercepts (where they connect with x and y axis, maximums and minimum peaks, asymptotes, etc.). Play around with some online graphing calculators like https://www.desmos.com/calculator or get a graphing calculator so you can see what functions look like if you don't know how to plot points and draw it. As you get into integral and multivariable calculus being able to draw graphs or at least visualize what they are is essential. Multivariable calculus especially because you're often calculating things like the volume of complex shapes. Integral calculus because you're doing things like calculating shapes as you rotate them around an axis to create a three-dimensional shape. In this problem you can draw the function and a tangent line that is a straight line that intersects your function at exactly one point. So, yes, in my opinion being able to draw it is extremely important.
not particularly. since you're looking for global max/min, you just find out how high the end points actually are, compare them to the height of the points where f'=0 and see what's highest and what's lowest.
as it turns out, f(0) actually isn't a global max or min, although f(3pi/2) is.
incidentally, technically the derivative isn't zero when x=3pi/2 despite what your derivative function might seem to suggest, since the derivative is not defined at the end point of an interval. Nevertheless x=3pi/2 turns out to be a global max anyway because it is an endpoint and therefore qualifies as a critical point for this reason.
check this: https://www.desmos.com/calculator/a60vuybaaw
If you're now claiming that the function is not a quotient after all, then of course this limit goes to zero. However where there is a quotient involved, involving Ce^(kx), there is no guarantee this goes to zero. See: https://www.desmos.com/calculator/heu4qb99x0
Not quite right. The derivative is -4x(x^2 - 4)^-3 . Then setting equal to zero, we see that x=0, and thus a max or min there. It's the only point, so it's absolute.
Use test points to see if it is a max or min. Try f'(-1) and f'(1). We can see the slope goes from negative to positive, thus the point is a minimum.
Plug zero into the original function to get the y value of the min point. f(0)=1/16
What should be noticed from the beginning is that x cannot equal -2 or 2 which would make the the function undefined and have vertical asymptotes at those x values.
Here is a graph of the function, and you can see the minimum at zero and the asymptotes at -2 and 2.
There are many online calculators for calculating the disk/washer method. Here are a few:
I just took Calc III over the summer. You’ve gotten a lot of good advice, but I’d like to add a few of the things I struggled with to see if they’re of help to you:
• If Calc II kicked your ass, I’d suggest you revise some subjects, ESPECIALLY if your weak point was integration. In Calc III, finding the bounds of integration can be very difficult sometimes, and it only gets complicated as you delve into triple integrals. On that note, make sure your integration techniques are spot on! Sometimes you can really simplify an integral if you use, for example, polars.
• Get comfortable with visualizing. Visualizing what’s going on and how elements within a problem intersect is really key to understanding and simplifying what’s happening. I personally used Geogrbra’s 3D calculator since it was allowed in my course, but try to develop the ability to do it on your own. It will help you down the road.
• Really dig into online resources. Sometimes one way of expressing a topic may not click for some people. Paul’s Online Resources is a great option for some, but I personally don’t do too well with it. I really like the resources that u/supersensei12 listed, so definitely give them a go.
To sum it up, revise your integration techniques. As one of my professor says: the key is knowing how to integrate. So long as your foundation is solid, you shouldn’t have too many issues. Good luck!
E is the region in 3D over which you'll be integrating this 4D equation. You are bounded between the x-values (or, if you prefer, the yz-planes at) 1 and 3, y-values (or, if you prefer, the xz-planes at) 0 and 2, and z-values (or, if you prefer, the xy-planes) 0 and 4. This is a rectangular box with lengths of 2 in the x-direction, 2 in the y direction, and 4 in the z direction.
See a graph of the region here: https://www.geogebra.org/3d/r2q7nu8h
I'm taking Calc 2 next semester in college and i still think Khan Academy is a great resource. If you're willing to pay a few bucks, Krista Kings courses on Udemy.com are really good.
For anyone taking Calculus 1
My brother has spent the last year working really hard to create a notebook specifically designed to be an all-in-one calculus 1 companion. It combines regular paper, graph paper and condensed, typed notes at the end for quick and easy referencing. Check it out on his instagram https://instagram.com/neatnotes_notebooks?igshid=9yylj1lq3i4u
Or on Amazon at https://www.amazon.com/Calculus-Notebook-Neat-Notes-Notebooks/dp/B08F2Y54NV
Imo don’t waste the money. Use tools for your phone/pc which will do a better job and stick to a scientific calculator. A lot of college courses won’t let you use graphing calculators on tests. So far I’ve had one class in two years who would allow one.
Casio has a ~$15 scientific calculator with natural display, nonCAS equation solvers, with solve vector math, systems, definite integration/differentiation, etc. great calculator allowed on pretty much all standardized testing. Also runs on solar and is smaller in size as an added bonus.
https://www.amazon.com/Casio-fx-115ES-Engineering-Scientific-Calculator/dp/B007W7SGLO/
I found nothing better than MIT OCW's 18.01 Full course here Videos are short 40-50 mins and very engaging, problem sets are challenging.
You should supplement this course with a traditional textbook, Thomas and Finney is a really good book, language is easy to understand and it doesn't have distracting bells and whistles.
Schaum's books are great, but I've never used the Calculus one specifically. I love all Dover's books. This calculus book has 4.5 stars, WTF. Dover, man. Some physics applications might be worthwhile to look over as well. I tried to find the book that I used when I went back to school but it's in storage. It was titled similar to "Calculus for Dummies" which, frankly those books are usually pretty good. That one might be worth it as well.
This is the book
https://www.amazon.com/Calculus-Early-Transcendentals-Howard-Anton-ebook/dp/B01DV7OEI2
Okay, it's only asking for part A of the problem, I was about to do it by graphing it out, but even then I won't really know where it wants to be integrated from looking at the graph.
Pick up pretty much any book on introductory analysis. I like this one. It will probably be more than you'll need, but it's only $10.
I'm in the UK and in college I did maths so it's A level mathematics which I have done. the book I needed to buy for this course is
Im sorry I'm not much help even to myself
I'll watch some khan academy videos currently trying to get my head around vectors. (I'm not sure what level but the questions go along the lines of defining the kernel of T, define the map T: P2(x)->P2(x) by T(p(x))=(1+x)dp/dx
stuff like that)
Do you know what part of vectors I should be learning to understand that?
Check around for a used copy, I picked up a used copy of Stewart 7th edition at the beginning of this semester off amazon for ~30 bucks.
Edit: try the used version here
http://www.amazon.com/Single-Variable-Calculus-Early-Transcendentals/dp/0538498676/ref=sr_1_2
I will probably get some flak for this but, Calculus The Easy Way http://www.amazon.com/dp/0812091418/ref=rdr_ext_tmb is the best way to learn calculus on your own. It is not an all comprehensive look at Calculus, but it will get you to understand the basics and give you a very good foundation for learning more. Best self learning, entertaining, way to learn Calculus hands down.
Glad youre getting back into the swing of things! Personally I'd recommend working your way through:
http://www.amazon.com/Calculus-Vol-One-Variable-Introduction-Algebra/dp/0471000051
Its excellent, will get you back up to speed, and provide you with the theoretical underpinnings necessary beyond Calc 3
If you can find a copy somewhere cheap.
http://www.amazon.com/Calculus-Transcendentals-Available-Titles-CourseMate/dp/0534465544
Also, check out your library to see what other calculus books they might have.