Could you help with this calculus question? (MVT)
Thanks for A2A, Paul.The statement is not valid in this form.As David Joyce showed, one possibility to repair it is to assume [math]ab>0.[/math] Counter-example: [math] a=-1, b=1, f(x)=x^2-1[/math]. Since [math]f(1)=f(-1)=0[/math], it satisfies the condition above.If there had been such [math] c[/math], it would have held [math] f(1) +f(-1) - (c^2-1) =(1 + (-1) -c) 2c[/math]So [math]c^2-1=2c^2[/math] and thus [math]c^2=-1[/math]. A contradiction.Moreover, I don't know what is a "Rolle function", I know only Rolle's theorem.
I need help with this calculus question, can anyone help?
A certain rational function f(x) contains quadratic functions in both its numerator and denominator. Aside from that, we also know the following things about f: f has a vertical asymptote at x = 5 f has a single x-intercept of x = 2 f is removably discontinuous at x = 1, since Based on these delightful clues, evaluate: (a) f(0) (b)
Can anyone help me understand how to do these calculus questions?
1. Can anyone explain Leibniz notation when finding a derivative? 2. How do you find vertical and horizontal asymptotes? 3. How do you find the second derivative? 4. What are and how do you use the 1st and 2nd derivative test? In this question: s(t)=t + t^-1, i found that the critical points are -1, 0, and 1. Are these right? How do you use those to figure out if it has a max/min or neither?What are points of inflection and how do you find them? What does it mean to be concave down/up? And how do you find the equation of a vertical/horizontal asymptote? PLease help! I have an exam and I don't understand this chapter at all!
Can anyone please help me with this calculus question =) ?
Cool... I like these problems: PART A) So the position of the ball is given. We know that the velocity is the derivative of the position equation: eq1) v(t) = ds/dt = 99.3 - 9.8t We also know that the ball is going to be at its highest when the velocity equals to zero. So plugging zero into equation 1. 0 = 99.3 - 9.8t t = 10.1 sec ---> that is when the ball is going to reach its highest height. Plug that number into the given equation and we get the answer to the first part of the problem. s(10.1) = 99.3(10.1) - 4.9(10.1)^2 = 503 meters. a) 503 meters. PART B) Okay, for part b we have to solve the position equation setting it equal to zero. We'll get two answers since it is a quadratic equation. One answer is before the ball is launched and the other is when the ball returns. So... S(t) = 0 = 99.3t - 4.9t^2 0 = (99.3 - 4.9t)*t So, t = 0 or t = 99.3/4.9 = 20.2 seconds There we have, answer for part b is 20.2 seconds. b) 20.2 seconds PART C Part C is easy, all we have to do is plug the answer from part B into the velocity equation that we derived previously in part A. v(20.2) = 99.3 - 9.8(20.2) = -98.66 m/s and it is negative because it is coming downward. PART D We know that acceleration is the derivative of the velocity. So we take the derivative of the velocity equation v(t) = 99.3 - 9.8t and we get a(t) = -9.8 m/s^2 which makes sense because there we know that the gravity is the only force pulling on it. It is not a function of time. So the answer to part d is that the acceleration is 9.8 m/s^2 at any given time. Alright....
Calculus problem need help from anyone thanks?
An object thrown vertically upward from the surface of a celestial body at a velocity of 18 m/s reaches a height of s= -0.3t^2 +18 meters in t seconds. a) Determine the velocity v of the object after t seconds. b) when does the object reach its highest point? c) what is the height of the object at the highest point? d) when does the object strike the ground? e) with what velocity does the object strike the ground? I need major help with this, if you can show it step by step with answers so I can follow along that would be great, thanks
I need help with calculus if any one can help, This is an example from the book, so i know the answers?
the equation has nothing to do with area , area is the function in which you want to Maximize. But the Restricting function is given by the perimeter of your fences. Here you have two adjacent pens, ( meaning two rectangles, that share one side. ) Looking something like this ( hopefully this comes out ) __ | | | ---- Where the base and the top are represented by y, and the 3 vertical lines are represented by x. so 3x + 2y = 100 ( all the fencing he has available. ) Then if you want to maximize your area, first solve this equation for one variable, and then plug that into your maximizing function. so you have y = ( 1/2 ) ( 100 - 3x ) Plug that into your area function. A = x ( 1/2 )( 100 - 3x ) Distribute that, then take the derivative, and solve it for when the derivative is zero, ( maximum's and minimums occur when your derivative is zero. ) I think you get the idea from here =)
Are online calculus help sites really helpful every time you need help with your calculus homework?
I wouldn't say every time, simply because sometimes I start to ask the question and the answer was that I need to check something and asking made the error obvious. (A large share of my errors are of the transposed-a-sign variety.)However, they are very helpful in general for a variety of reasons, including the fact that you can get explanations in a lot of different styles, for a lot of different learners. This can be very useful.Some students also have the idea that they should be fast and not need help, and can find asking someone else for help to be embarrassing--websites can answer your questions without anyone else having to be involved, and they don't care how long you linger on a page.I happen to really like Paul's Calculus Notes, since his notes include explanations and example problems, and I find his explanations very intuitive. Wolfram Alpha can also be helpful--I primarily use it to check my answers, since I don't have a pro account and don't get step-by-step answers.While this isn't a calculus site in particular, I also recommend Proofapedia. It's handy.
Can anyone help me with pre-calculus functions problems?
Ok, these questions are all fairly similar, so I'm not going to go over each one individually.Several are giving you varients of x^2. Normally, this function isn't one to one. In this case, that means there is more than one input that gives the same output (e.g. -2^2 and 2^2 both output 4). In order to make the function one-to-one, all we have to do is say x≥0 OR x≤0. So we either say that we can only have positive imputs and 0(which is probably the answer they want) or we can only have negative inputs and 0.What if it's something like 4x^2 instead? Just set 4x^2≥0 and solve the inequality. The answer should always be in terms of x, not 4x^2. That's why you need to isolate x in the inequality.You then are supposed to restrict the domain of the inverse function to make it one-to-one. The squareroot function normally refers only to the positive sqrts. This is what they want, not the negative sqrt function (neither is really more correct, but it's convention to go with the former).They also most likely want you to restrict the domain so that you take the square root of a positive (sqrts of negatives are defined, but they aren't part of the real number system they want you to work with). So, for example, if you have the function 2x^2-10, to find the inverse function you need to set 2x^2-10≥0 and solve for x. 2x²≥10 ⇒ x²≥5 ⇒ x≥√(5). This would make your inverse function √(2x²-10) where x≥√(5).You also have problems with composite functions. That's what f circle g means (i.e. f(g(x)). That might look confusing but it's actually fairly simple. Say f(x)=x+5 and g(x)=x³ f(g(x)) would be x³+5. So if for those problems, just get your composite function and then proceed as normal.I hope this helps! Feel free to ask follow-up questions in the comments. :)Also, check out these websites. They are very helpful when learning math:Math is Fun, Khan Academy, Math lessons that click, Purplemath | Home
Can anyone help me with this calculus problem?
a) Profit = income - cost = 800x - (500 + 0.02x + 0.001x^2) P(x) = 800x - 500 - 0.02x - 0.001x^2 To maximize, take the derivative. P'(x) = 800 -0.02 - 0.002x Set derivative = 0 to find max and min points. 800 -0.02 - 0.002x = 0 x = 399990 And since the equation P(x) had a - value for the x^2 term, the parabola is /\ shaped, so the one critical point is a maximum. Therefore, the rate 399990 mills per day is the rate of production that will maximize P(x), profit. b) Maximum daily profit is just the value of P(x) at the value of x found in part a. P(399990) = $159991500.10
What are some of the best calculus books for IIT-JEE?
When I was in Class XII, I was absolutely crazy about Calculus, after all it was the field which attracted me to study Math for my higher secondary level and above. I was in constant search of good study materials, books. If you have a command over NCERT, then you can go for R. D. Sharma, Course in Mathematics for IIT-JEE (TMH), M. L. Khanna. I. A. Maron is definitely a recommended book, many early problems in IIT-JEE (1970's, 80's) can be found in this book. I also heard of Thomas/Finney but never got a chance to read it. I came across a very good book called Teach Yourself Differential/Integral Calculus but the book was not aimed at IIT-JEE. I would also recommend K. C. Sinha as I cleared my many concepts during B. Tech. by studying from this book. In recent times, I have seen the following book that seems strikingly good: Introduction to Differential/Integral Calculus: Systematic Studies with Engineering Applications for BeginnersAuthor(s):Ulrich L Rohde; , G. C. Jain, Ajay K. Poddar, A. K. Ghosh.All The Best!!!(P. S. I didn't clear JEE, but keep myself updated, tutored a few students for Class 12 and entrance)