what is the relationship between f(x) and f[prime](x)?
355882
The relationship is defined in one dimension as f'(x) being the slope of the tangent line to the graph of f(x) at the point x. In symbols, f'(x) is the limit as h approaches 0 of the slope of a secant to the graph passing through x: [f(x + h) - f(x)]/h.\nf'(x) being the slope of the tangent line to the graph of f at x has its advantages. When f'(x) is 0, we know the tangent line is horizontal there, meaning the graph is either at a minimum, a maximum, or a temporary flat area. Higher derivatives (treating the derivative as a function itself) can tell us which of the above it is.\nDerivatives are most concrete in physical examples. If x(t) is the position of an object at time t, then the velocity of the object at time t can be idealized as the derivative of x(t) at t (instead of the average velocity, which is the slope of a secant line to the graph of x(t)). Similarly, the acceleration at time t is given by the second derivative.\nIn higher dimensions, where x = (x1, x2, ..., xn) is a point and f(x) = (f1(x), f2(x), ..., fn(x)) is also a point, this is generalized to f'(x) being the unique linear function that vanishes with the change in f at x. In symbols, the limit as ||h|| vanishes of [f(x + h) - f(x) - f'(h)]/||h|| vanishes. This coincides with the definition in 1 dimension above. In higher dimensions, f'(x) corresponds to the basis of a tangent space at x, as it did in 1 dimension.\nIf f is a map from R^2 into R, f'(x) is a linear map from R^2 into R, which is a plane. This plane is a tangent space to the graph of f at x. Higher-dimensional tangent spaces are not as visualizable.
904043
One variable is related to another by a function involving constant terms and terms of 1st order or higher.
377581
I am going to assume that you are talking about the following:\n n\n ∑ f(i)\ni=k\nThis means the sum of f(i) for i=k to n.  In other words,  that expression equals f(k)+f(k+1)+f(k+2)+...+f(n-1)+f(n).\n\nThe "pi notation" that I think you must be talking about is this:\n\n n\n Π f(i)\ni=k\nThis means the product of all of the f(i)'s from i=k to n.  In other words, this expression equals:\nf(k)*f(k+1)*...*f(n).\nClearly very similar to the sigma notation.
517319
Using the definition of the derivative.\nf'(x) = lim(as h->0) of [f(x+h)-f(x)]/h\n\nLet f(x) = mx + b\nf(x+h) = m(x+h) + b\n\nf(x+h) - f(x)\n= m(x+h) + b - (mx+b)\n= mx + mh + b - mx - b\n= mh\n\n[f(x+h)-f(x)]/h \n= mh/h\n= m\nTherefore,\nf'(x) = lim(as h->0) of [f(x+h)-f(x)]/h\n= lim(as h->0) of m\n= m
311134
f(x) = 1/(x+5)(x+2)\n\nso -5 and -2 are restricted from the domain since then we would have the denominator = 0 and the function undefined.
499879
I'm trying to guess what the problems you were asked to do were, because the information you gave in posing the question was incomplete.\n\nIf f(x) = log(x)  (for log of any base)\n\nthen\n\nf(1/b) = log(1/b) = log(b^(-1))=-log(b)=-f(b)\n\n\nlog(MN^2) = log(M) + 2 log(N)\n\nor\n\nlog((MN)^2) = log(M^2) + log (N^2) = 2log(M) + 2log(N)\n\nThese are applications of the laws of logarithms.
1005272
f(4) means 'rewrite f(x) = 2x - 5 with a "4" instead of an "x"', like this:\n\nf(4) = (2 * 4) - 5\n=8 - 5\n=3
760867
f(x) = (x+3)(x^2 + 16)
464622
Use any recursive algorithm.\n\nFor instance:\n\nf(x) = f(x) + 1
977406
Here is an outline:\n\nf is an ONTO function. So for EVERY element b in B there is an\nelement a in A so that f(a) = b.  That is what onto means.  All the\nelements in B have some element in A that f maps it into.\n\nSince g is not one to one there must be two elements  in B that get mapped into the same element in C.  Call them b1 and b2. Show that b1 and b2 are part of the codomain of f.  f(x) maps into all B, so there must be a a1 and a2 that get mapped into b1 and b2.  So then we have g(f(a1)) and g(f(a2) that map into the same element, and hence g(f(x)) is not one to one.
741312
If f(t)=(t-a)*(t-b) = t^2-(a+b)*t+ab; then \nI=(1/3)*t^3 - (a/2+b/2)*t^2 + a*b*t+C
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.
530638
only could work with known answeres.\nassuming f== pt i,2 , f*==-1.-2, x=0, mid point.\ndf*-1/2dx(2)= df, because the equation blances out the negotive and positve,\nf(x) =x*5 + 2x -1,x as 0. f(0) = o*8+2(0) - 1 answer points are -i,1theni culdn't figure out how to find the answers for line y,x. or the solpes\nthe last one i tried was\ndf*-1/dx (2)  i got =0 df = points 1,2 + -1/2 cancels out the equation.\n\ni thnk, maybe, might be so? could you e-mail me the correct answer?
695595
Each question is a function of two variables... and it has two derivatives, relative to each variable.\n\nDerivative of F relative to Z1   is   uZ1^(u-1)*Z2^v\nDerivative of F relative to Z2   is   Z1^u*vZ2^(v-1)\n\nDerivative of MRTS relative to Z1 is  -uZ2/(vZ1^2)\n\nDerivative of MRTS relative to Z2 is  u/(vZ1)
795057
First show that the reciprocal of every element of F[[x]] can be written as a Laurent series: this is best done by inductively defining the coefficients of the quotient. Conversely, show that the reciprocal of every Laurent series can be expressed as a Laurent series. This shows that F((x)) is a field (assuming you check the obvious properties) and contains the field of fractions of F[[x]]. Since a sufficiently high power of x can be multiplied by a Laurent series to get an element of F[[x]], this completes the proof that F((x))) is the field of fractions of F[[x]].\n\nAs for Z[[x]]. The idea is to show that the power series for e^x (which is clearly in Q[[x]]) is not of the form f(x)/g(x) for f,g in Z[[x]]. But if f/g is in lowest terms, it is easy to see that g must be a constant (otherwise there would be terms of the expansion with negative exponents), i.e. g is in Z. But the expression for e^x cannot be so written!
794664
1) If h(x) = f(g(x)) = (x^3 -1)/(x^3 +1), then \nf(x) = (x-1)/(x+1) and g(x) = x^3\n2) If h(x) = f(g(x)) = 3sqrt(x^2 -8), then\nf(x) = 3sqrt(x) and g(x) = x^2 -8
883154
I assume that's suppose to be an integral symbol.\n\nf(x) = [1/2 t²] from 2 to x, or 1/2 x² - 1\n\nThen f'(x) = x, and f''(x) = 1.\n\nDidn't really need to do the integration...