The Mean Value Theorem?
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The mean value theorem says that if a function is continuous and differentiable on an interval (in other words, there are no breaks in the function and it is a smooth curve, without any "cusps" which are the pointy bends), then somewhere on the interval, the value of the function will be equal to the average value found by evaluating the end points then dividing by the length of the interval. That's what (f(a)-f(b))/(a-b) is -- the average or mean value. \n\nI hope that helps.
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When you say stationary value, do you mean that the rate of change is zero? That is, where the slope of the line tangent to the function is equal to zero?\n\nIf so, then yes, at least conditionally.  When the slope of the tangent line, or the rate of change, is equal to zero, then the value of the function at that point is what we like to call a "local maximum" or "local minimum."  This means that, in the immediate area, or for values of the function close-by, that this point is the maximum or minimum.  However, these places on a function may or may not be what we like to call the "absolute maximum" or "absolute minimum."  There may be other values for the funtion that are greater than or less than the stationary value you are talking about.  If you are taking Calculus, then you should learn more about this.  If not, then this is pretty much the most broad explanation I can make without talking about Calculus.
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It depends on what the word "mean" means. :-)\n\nRoughly speaking, under the specified conditions there is a point on the curve between x=a and x=b where the slope of the curve equals the slope of the secant from (a,f(a)) to (b,f(b)).
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No. Look at f(x) = |x|. It is everywhere continuous but not differentiable with respect to x at x=0.\n\nThis is a stupid question, and you're stupid for asking it.
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Check out the Bernoulli principle (try wikipedia).  That will explain all.
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what was the question?\n\n\nDo you want to know terms like mean, average, mode, standard deviation, etc.?
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It takes only one example to prove the statement to be false, and I provided one:\n\n(Now, while you may never naturally encounter the set of numbers I used, the definitions of mean and median fit, so the statement is false.)
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You might try reading Whitehead and Russell's Principia Mathematica.  I can't promise it will help, as I haven't read it, but I've been told that this issue is addressed.
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I wont give you the entire proof, just give you the key arguments.\n\nknowing the joint distibution of (X, Y) allows to compute the marginal densities of X and Y.\nIndeed, let f_X be the marginal density of X, then:\n\nf_X(x) = &#92;int f(x, y) d_y\n\nf_Y(y) = &#92;int f(x, y) d_x\n\nNow, you might compute E(Y| X = x) = &#92;int y f_{Y|x} (y) d_y\n\nwhere f_{Y| x} is the conditionnal density of Y knowing the event X = x\n\nrecall that f_{Y| x}  = f(x, y) / f_X(x), and you just have to compute the resulting integrals\n\nregards,\n\njfp
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Try,\n\nhttp://www.personal.psu.edu/faculty/d/x/dxh40/MBLemma.pdf
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Squeeze theorem:\n\nIf you know that something is always between to points (inclusively) and that those two points eventually come together (or their limit comes together), then you know that the point is equal to the limits of the others.\n\nFor your example:\n\nsin(u)≤1 for all u.  Thus sin(1/x)≤1 for all x≠0\nLikewise sin(1/x)≥-1 for all x≠0.\n\nThus -x≤xsin(1/x)≤x for all x>0.  Since -x-->0 as x--->0+ and x--->0 as x-->0+, by the squeeze theorem xsin(1/x)-->0 as x-->0+\n\nThis can be repeated to show that xsin(1/x)-->0 as x-->0- also, and thus the right and left limits equal and the limit is zero.
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You're given the equation, silly person.  You absolutely need it to find the mean of a sequence.\n\nIf you can't find it, check out this website.
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Counterexample.  Let u(x), v(x) be members of {sin x, cos x}.  Both of those are solutions, but you do not get zero by adding one to the other, or by doubling one of them.
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Here is a link you may find helpful.\n\nhttp://www.math.buffalo.edu/mad/00.INDEXmad.html
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There is a local maximum at x = c on f(x) if\nf'(c) = 0 AND f"(c) < 0\n\nOccasionally also when f"(c) = zero .. this occurs when there is a multiple root at that value of x.\nThen further testing is needed. In this case:\n1. if If f(c-) < f(c) AND f(c+)< f(c) then it is a local maximum;OR\n2. if f'(c-) >0 and f'(c+) <0 then it is a local maximum; OR\n3. if f"(c-) <0 and f"(c+) <0 then it is a local maximum
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The integral of a function over a definite interval.
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If the you have a local max/min at an interior point, , the partial derivatives must be zero.\nSince the sum of the second partials is zero, the point must be a saddle point, not a max/min.