Is this calculus proof valid?
920889
if f is differentiable in x+0 then its also continue in x_0 :\n\nf(x) - f(x_0) = f'(x_0)(x-x_0)+ E(x)\n\nlim x->x_0 then left side 0 and right side 0 thus f(x) = f(x_0) for x->x_0\nthus continous in x_0
64159
It's debatable.  Here's an interesting column that addresses the question:\nhttp://mathforum.org/library/drmath/view/57188.html
344009
No, because it doesn't.\nThere's a number of "proofs" out there that rely on dividing by 0, even when it's not obvious that's what's going on.\nDivide by zero, and you can prove anything!
837082
Hint: Let C=X&#92;O and let f(x)=dist(x,C). The other way is trivial.
487867
Here is the mathematical proof:\n\nIt involves some precalculus, so I will explain what I am doing just in case you haven't gotten to precalc yet.\n\n[1]  Definition: Limit\n     The limit of a function f(x) is the value of y @ x.\n     i.e. y = 2x^2, @ x = 1, y = 2, \n     Therefore, \n     limit 2x^2 = 2\n     x->1\n\n[2]  Why division by Zero = infinity:\n\n     limit 1/x = infinity\n     x -> +0\n\n     Limit 1/x = infinity\n     x -> -0\n\nThe graph of y = 1/x explains this principle. \n\n[2.1] Definition: +0  & -0\n\n+0 = As x approaches zero from the right.\n-0 = As x approaches zero from the left.\n\n[2.2] As x get closer and closer to zero, the quotient of 1/x gets larger and larger.\n\ni.e. 1/0.0000006 = 6000000\ni.e  1/1x10^-9999= 1x10^9999\n\n[2.3] When x becomes zero, the quotient of 1/x is infinity. Since infinity is underfined, 1/0 is also undefined!
765864
I'm unclear with the way you stated the problem. Am I looking for a implicit derivative. I'll check back to solve it.
619240
u can even use calculus to find this out
779791
(f(2) - f(0))/(2 - 0) = 2, but f'(x) < 0 for every x, except x=1 where it is undefined.\n\nThe function has to be continuous in [0,2] and differentiable in (0,2).  For this f(x), it is undefined at x=1, so the conditions of the MVT are not met.  Math is still safe!\n\nCheers!
830837
If ∫f(x)dx = y(x), then f(x) = d y(x) / dx;  So to get the answer, take the derivative of the expression x^2 - 2x + 1.  The value of the integral at limit 1 is  a constant and will not appear in the derivative.
632327
Write f(x,y) as x^2*[y^2/(x^2+y^2)]. The second factor is always <=1, so f(x,y) goes to 0 as (x,y) goes to (0,0). The first answerer is WRONG. As for the second answerer, just knowing the limit is 0 along every line is not enough to conclude that the limit exists. It is possible for limits along lines to be 0, but to be otherwise along parabolas.\n\nIf f(0,0) is defined to be 0, then this function has partial derivatives at (0,0) that are both 0. In fact, f is differentiable at (0,0) in that case. The derivative is the 0 functional.
278447
answer No 1 is bullshit. Answer No 2, I couldn't even understand what he/she wants to say.\nBut answer 3 is correct:\nint(e^(x-y)=int(e^-y e^x) = e^-y int(e^x) = e^-y e^x +c=\ne^(x-y) +c
914585
f(x) satisfies the mean value theorem because it is continuous and differentiable on [-2, 1].\nSecant slope = (1 - 4) / (1 + 2) = -1\nDerivative f'(x) = 2x\n2x = -1\nx=-1/2 (this is what c equals to)
408554
yes. take out an x \nyou answer should be x(x+3)
932028
Yes.\n\nThe argument is valid but not perfectly worded.  \n\nHere A, B, C can be variables as others have pointed out or they are statements that are fulfilled in direct progression.  In fact this is a form of the "Direct Proof"\n\nDirect Proof example:\n\nAs an example we can claim that if we have A:={the addition of two odd numbers x, y |  x+y} it will become an even number which can be statement C.\n\nSo by our knowledge that odd numbers can be written as \nx = 2m+1   and y = 2n+1  where m, n are any integers we can use that as statement B.  Hence B:={x = 2m+1   and y = 2n+1  where m, n are any integers}\n\nDoing a little algebra we have \nx+y = (2m+1) + (2n+1) = 2m + 2n + 2 = 2(m+n+1)\n\nNow we can find out that (m+n+1) is an integer as by definition and we can call it "p" s.t. p=(m+n+1),\nHence x+y = 2(m+n+1) = 2p, and by our knowledge of numbers 2p is an even number.  Hence C is fullfilled.\n\nThus like questioner's argument:\nIf A then B then C\nA \ntherefore C.\n\nAnd with my example and statements in quotations can be be replaced by letters A, B, and C: \nIf "you add two odd numbers" then "they are integers that can be written as the addition of two times another integer plus one which algebraically can be written as two times integer" then "you obtain a number which is even".\n"You add two odd numbers" \ntherefore "you obtain a number which is even".\n                                              Q.E.D.
937153
B is false: Let a be any number other than zero. If a>0, 2(2a)-a > 0, and f'(2a) = 2. Conversely, if a<0, 2(2a)-a<0, and f'(2a) = -2. Thus, any value of a other than zero serves as a counterexample.\n\nA is also false, although I'm not sure your teacher knows this. She may be thinking of something like f(x)=[a, b]∫g(t) dt, which is alawys a constant function regardless of g(t), and thus its derivative is zero. However, if the free variable appears in the limits of integration, the function may be nonconstant: for instance, let f(x) = [1, x]∫1/t dt, which is equal to ln x for positive values of x, and its derivative is certainly not zero. So the correct answer here is d: both A and B are false.
942804
I'm no calculus genius or anything but...\n∫dx/(sec^2)(5x)\n∫(cos^2)(5x)dx\n∫(1+cos(2*5x))/2 dx\n∫1/2dx + ∫ cos(10x) dx\nx/2 + sin(10x)/20
875283
You differentiated and found it has relative min at x=1/3?  but if you differentiate again you find second deriv is\n54x - 1/(x^2), and I think this goes from \nnegative when x<1/cuberoot(54) to positive when it's greater, so that means it has an inflection.  Also means it's concave down when \nx<1/cuberoot(54), so the answer must be 5, not 2.  Unless I made a mistake!\n\nPS  Dave is wrong about 1.  Domain is all positive real numbers, because lnx is undefined for 0 and neg.