Please can anyone help with continuity and differentiability calculus problems?
819945
1.  the idea here is to take the limit for both pieces and show that they match.  This is relatively straight forward.  The limit as x approaches 1 from the left of x cubed is simply "1"     \nThe limit as x approaches 1 from the right of 3x-2 is 3(1)-2 also equals 1.  Since these two match, we have proven that "f is continuous at 1"\n\n2. The fact that f is continuous at 1 is INsufficient to imply that is is differentiable there\n\n3. the derivative from the right is simple 3 since the derivative of 3x-2 is 3.  The derivative from the left of 1 is also 3 since:\n\nf(x) = x^3 implies that f'(x) = 3x^2 and f'(1) = 3(1)^2 = 3 as well.  Since the derivative from the right equals teh one from teh left, the function is differntiable at x =1
99672
honestly....good luck\n\ntry www.mathworld.com\n\ndo a search for differential equations, i think that would be your best bet...but that can get confusing...good luck
921398
I just taught this section today to my calc students.  You start by finding the first and second derivative.  I recommend looking in your book for an example.  Your book probably has something like this towards the front of the section.  It would be too long winded to write everything down here.  If you don't have your book on you, the web page below has problems that you can do and then check your answer.\n\nhttp://www.math.ucdavis.edu/~kouba/CalcOneDIRECTORY/graphingdirectory/Graphing.html
664515
There is a defined term oscillation used to describe how much a function fails to be continuous at a point by evaluating the difference between the supremum and infimum of values of f(x) in some non-zero neighborhood of x as the radius of that neighborhood goes to zero. The most this can do is show that f is continuous whenever f(x) = 0, though, and continuity obviously does not imply differentiability (See the absolute value function). I'm guessing that you're therefore just referring to mathematician's interpretation: the tangent line oscillates infinitely near the origin, leaving no derivative.
813038
f(x) = x^2 / (x - 2 ) ^2?\n\n    (a) Find the vertical and horizontal asymtotes\nx=2\n    (b) FInd the intervals of increase and decrease\nfind f'(x) = ( (x - 2 ) ^2 2x - x^2 2(x-2) )/(x - 2 ) ^4\n= ( x-2) ( (x-2)2x-2x^2) / (x - 2 ) ^4\n= ( 2x^2 -4x -2x^2 ) / (x - 2 ) ^3\n= -4x/ (x - 2 ) ^3  \n=0\nif -4x =0, i.e. x=0 one critical point.\n    (c) Find the local and maximum and minimum values\n    (d) Find the intervals of concavity and the inflection points
927535
#1 \n\nIf f(x)= √(x³ + 5x +121) * (x² +x +11) \n\nthen, f(0)= √(121) * (11) \n= 11 * 11\n= 121\n\nf'(0) = -1/2 (See #3 below at ***)\n\n#2 \nIf y is a differentiable function of x, then the slope of the curve xy^2 -2y + 4y^3 = 6 at the point where y = 1 is what?\n\nWhen y = 1\n\nx(1)^2 - 2(1) + 4(1)^3 = 6 \nie x + 2 = 6\nSo x = 4\nSo y^2 + 2xydy/dx -2dy/dx + 12y^2dy/dx = 0\nSo 1 + 12 dy/dx - 2dy/dx + 12 dy/dx = 0\nThus 22dy/dx = -1\ndy/dx = -1/22\nie slope of tangent at (1, 6) is -1/22\n\n#3\nFind all open intervals on which the following function is decreasing:\nf(x) = x / (x^2 + x -2)\n= x/[(x + 2)(x - 1)]\n= 2/[3(x + 2)] + 1/[3(x - 1)] (by partial fractions)\nDiscontinuities at x = -2 and x = 1\n\nf'(x) = -2/[3(x + 2)²] - 1/[3(x - 1)²]\n\nSo f'(0) = -2/[3(2)²] - 1/[3(-1)²]\n= -2/12 - 1/3\n= -1/2 ***\n\nClearly f'(x) < 0 for all x except x = -2 and x = 1 where it is undefined \n\nie the function is decreasing fo x< -2, -2 < x < 1 and x > 1
754733
Let y = u/v   where u=some function and v =some funtion\n\nthen dy/dx = (v*du/dx - u*dv/dx)/ (v^2)     **Quotient Rule**\n\nIn this case u = 6x^2-1 and v = 2*sqrt(x)\n\nSo you just differentiate and plug into the equation for dy/dx
493069
First :\nSince f is continuous over its domain (R),\ndf/dx = 5 x^4 +2 > 0 for every x\nSo f is an increasing function over its domain\nHence for every value of y their exist a unique value of x\nHence f has an inverse and one-one function\n3- Use function graph to graph f-1 as you mentioned\n(df^-1/dx) (2)= the slope of the tangent to f-1 at x=2\n6- (1,2)\n7- 7\n8- 1/7
262508
Differential and Integral Calculus\nTopology of the real line\nLinear algebra\nAbstract algebra\nComplex Analysis\nDifferential Equations
888343
Use the chain rule:  dy/dx = d/dx[(ln x)^x] = x(ln x)^(x-1)d/dx[ln x]
766685
You need to differentiate the equation y=2x^2. \n\ndy/dx = 4x therfore the gradient at x=3 = 4*x = 12
475920
Homework?\n\nSuppose f is not identically 0. Then there is an interval where f stays away from 0. The g be the step function of this interval to get a contradiction.
1001167
The solution to your question is a bit lengthy and it wont be easy typing that here. However I can tell you the guidelines to arrive at that.\n(1)Write down the equation\n(2) take differentiation wrt x on both sides.RHS will be 1 here.LHS will have to be diff..you must know how to diff e^(y/x).Use the diff rule for products.\n(3) take diff.again of the above obtained equation.you will get alot of terms then.try getting from that the form you want.
118732
good luck man... http://mathforum.org/library/drmath/sets/high_calculus.html
1031874
There are a few requirements you need to fulfill to make f(x) differentiable everywhere:\n1)  The function needs to be continuous at x=2\n2)  The limit of f(2)-  [the limit of f(x) as you approach x=2 from the left] must equal the limit of f(2)+ [the limit of f(x) as you approach x=2 from the right]\n3)  f'(x) at x=2 must equal the same number -- for both x^2 and mx+b.\n\nSo, let's solve it out.\n\n1)  To be continuous, 2^2 = 2m + b.  So 4 = 2m+b.  So b = 4-2m.\n2)  We'll check this last.\n3)  When f(x) = x^2, f'(x) = 2x.  When f(x) = mx+b, f'(x)=m.  So 2x=m.  So let 2*2 = m.  So m = 4.\n\nGoing back to (1), we have b = 4 - 2(4) ==> b=-4.\n\nLet's check (2) now.  The limit of f(2) as you approach x=2 from the left is x^2, or 4.  The limit of f(2) as you approach x=2 from the right is 4(2)-4, or 4.  The 2 limits are equal, so we're ok.\n\nANSWER:  m=4, b=-4.
120626
There's a derivatives tutorial on the following page:\n\nhttp://www.karlscalculus.org/calculus.html#s4
1122081
It's called the integration by parts formula.\nInt u dv=uv- int v du\n\nBoth u and v need to be absolutely continuous for this to work. A continous derivative for both is plenty.