If  f  is a non constant entire  function then show range of f is dense  in  complex?
442777
Suppose that the range of f does not intersect some neighborhood of w. Show that 1/(f-w) is a bounded entire function. Louiville will finish it off.
445923
The domain of a polynomial function is all reals.\n\nThe domain of a rational function is limited by the domains of its numerator and denominator, and the denominator cannot equal 0. \n\nSince this is a rational function of polynomial functions, the domain could only be restricted by having the denominator being equal to 0. But the range of the denominator is y >= 3. Since there is no value of of x that can make the denominator equal to 0, the domain of the entire function is all reals.
491386
Just substitute x+1 into f(x)\n\nf(x+1) = [ 6(x+1)^3 ] - [ 3(x+1)^2 ] + 4(x+1) - 9
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
1051867
The range is:\n(-1)<x<=(1)\nx=1 is admissible
388327
I'm not sure what you mean by "Liavally" but I'll prove this using the definition of being analytic in a region.\n\nWrite f(x+iy) = u(x,y) + i v(x,y) in terms of real functions u and v. Ths function f(z) is analytic if and only if u and v satisfy the Cauchy-Riemann equations:\n\n{partial u}/{partial x} = {partial v}/{partial y}\n{partial u}/{partial y} = -{partial v}/{partial x}\n\nNow since f(z) >= 0 we must have v(x,y) := 0.  Hence {partial u}/{partial x} = {partial u}/{partial y} = 0 so that u(x,y) is a (nonnegative) constant.  This shows that f(z) must identically be a (nonnegative) constant.
207423
Not the answer, but a suggestion.\n\nThis is a college level questionm, so I'm assuming that you are in college. Go to the math department library and find other texts on complex analysis. Look at the chapters that cover the same material. Look at the examples the authors have worked out. You may be surprised to find this problem or one enough like it to give you a clue on how to proceed.\n\nThis is not that tough a problem. N years ago, (where 0 < N  < 50) I could have given you an answer, but since then, the analysis sections of my brain have shrunken.
546449
define f(x)=d(x,A)/[d(x,A)+d(x,B)].\nThis is a continuous function f:X->[0,1] (use triangle inequality to show x->d(x,A) is continuous for a closed set A). Also f=0 on A and f=1 on B. Let C=f^(-1)[0,1/3) and D=f^(-1)(2/3,1]. Then C and D are open because inverse images of open sets under continuous functions are open.\n\nNote to Adolfo Boy: It is possible for the distance between two disjoint closed sets to be 0. Consider the x-axis and the y=1/x curve. Your argument does work for compact sets, though. The point (a,b) in your argument need not exist if the infimum is not attained.
364736
This statement can be proven by proving the contrapositive: If f:R -> R is neither strictly increasing nor strictly decreasing, then f is not a 1-1 differentiable function.\nIf f is differentiable, then f being neither strictly decreasing nor strictly increasing is equivalent to f being a constant function over some nondegenerate interval in its domain. Thus f cannot be 1-1.\nIf f is 1-1, then the condition implies that f is not continuous, and thus is not differentiable.
121355
well if your taking the derivative, then there should be no h's left over at the end, since they go to zero with the limit.  The derivative of that function is [5(5x+2)+25x]/(5x+2)^2.  I'll leave it to you to simplify.
820313
If f : A-->B is surjective map, then its onto but not one-one .\nso you can write domain of f as set of all values that would give values in B.\n\nYou can write as\ndom(f) = {x : f(x)=y, x is in A, y is in B}\n\nSince the function is not one-one, there can be more than one x giving the same values of y, I dont think you need more conditions in set representation.
918808
wouldnt it be like 3<x<-3 f(x)\nso like x would be 1 so it would be just f if you really want to simply that far done.
863364
Not sure if I understood your question. maybe you meant a series of functions. Suppose that, for each n you have a function f_n and, for every x in the domain of the functions f_n, you define S(x) = f_1(x) + ....f_n(x) +...provided this series converges. Then, you have a series of functions, since for every x S(x) is given by the limi of a series.\n\nFor example, consider s(x) = 1 + x  x^2/(2!) ....+ x^n(n!)....|This is a power series, in this case it converges for every real x. This is the definitioj of the exponencial function e^x
401175
note that f^3/f^2 = f except for the points where f(z) = 0 (thos will be singularities).  But you can show that those singularities are removable (that will be your job - just remember that the order of the zero of f^3 will be larger than the order of the zero of f^2...)  Since they are removable, you can extend this to an entire function.  But, this function must be f, since they agree on a set with a limit point (namely, the set of f(z) s.t f(z) is not zero)... We should note that none of this will work if f is constant, but the constant case should be easy, right?
760867
f(x) = (x+3)(x^2 + 16)
779960
This is true of any distribution that has a density, not just the normal one. It follows straight from the definitions. The c.d.f. (cumulative distribution function) of a vble X is by def. the fctn F(a) = Prob(X < a). Density, g, is the derivative of F: g(a) is by def. the limit of (1/epsilon)Prob(a < X < a+epsilon) as epsilon decreases to zero. So F(a) is the integral of g from -infty to a (this uses F(-infty) = 0 and the Fundamental Thm of Calculus, that integration reverses differentiation).
786224
Actually, kaksi_guy seems to be onto something. Unfortunately, not only does his proof have the flaw that you noticed, but even if that were rectified, he would still have only proved that f is zero at infinitely many points on the interval, not that f is zero on every point in the interval. Still, his post is the inspiration for what follows:\n\nFirst, we need to assume that g(x) is defined everywhere on the interval between the two points. We don't need to assume that g(x) is continuous, but if we allow a function that is undefined at even one point, such as g(x)=(x-3/x)/2 (which is undefined at zero), we may easily generate exceptions to the theorem (in this case, f(x)=x²-1).\n\nWith that out of the way, we can begin our proof. Suppose f(x) = 0 at two points a and b. By the extreme value theorem, f(x) has both a maximum and a minimum on the closed interval [a, b]. Let us denote the minimum value f(x) attains on [a, b] as min(f(x)) (and analogously for max(f(x))). Now, since f(a)=f(b)=0, min(f(x))≤0 and max(f(x))≥0. Suppose that min(f(x))<0. Then the minimum does not occur on the endpoints, so it must occur on a point c in the open interval (a, b). It is readily shown that this implies f'(c)=0. Therefore, from the original differential equation, f''(c) - f(c) = 0 and thus f''(c) = f(c). However, f(c)<0, so this implies the function is concave down at c, contradicting the fact that c is a minimum. Thus min(f(x))=0. Similarly, if max(f(x))>0, then there would be a point c on (a, b) which is a maximum, but is concave up, which is also a contradiction. Thus, min(f(x))=max(f(x))=0, and a function whose minimum and maximum are both the same is a constant function, in this case f(x)=0 on the interval [a, b]. Q.E.D.