f;(0,1)-R by f(x)=0 if x in irrational =1/n if x=m/n.g:R-R by g(x)=0 if (x<=0 or x>1/2 )   =1 else.Show gf is not R integrable
137713
Well, if I understand your question correctly, whether gf is R integrable is the same as asking if f is integrable over (0,.5].  Then, the issue becomes that if you do a Riemann integral (ie., the sum of area of increasingly thin rectangles under the function but above the x-axis), the area chosen to represent any given single slice in a Riemann sum will be either its width times 1/n or 0 depending upon whether we pick a rational x or not at which to evaluate our integrand in that interval.  As a result, there is no limit to the sum of the rectangles.
260494
Let f(x) = 1 / (1-x) = 1 + x + x^2 + x^3 + x^4 + ...\n\nThen f'(x) = 1 + 2x + 3x^2 + 4x^3 + ...\n\nCompare that to y and you see y = x f'(x).\n\nSince f(x) = 1 / (1-x) = (1 - x)^-1,\nthen using (d/dx)(u^n) = n u^(n-1) (du/dx) you get\nf'(x) = -1 (1 - x)^-2 (-1) = 1 / (1 - x)^2\n\nThus y(x) = x / (1 - x)^2, and so\ny(1/2) = (1/2) / (1 - (1/2))^2 = (1/2) / ((1/2)^2) = 2
73250
Assuming you mean over R, solve the equation s*t + r*sin(t) = 0 for r and s for all t that is not the trivial solution (r,s)=(0,0). \nIn other words, if f(t)=sin(t) and g(t)=t were linearly dependent, then there is some non-zero slope s, s*g(t) = s*t and some non-zero amplitude r, r*f(t) = r*sin(t) such that the graphs of these functions completely coincide for all t. Since f is non-linear and g is, this is nonsense and the functions are linearly independent.
432505
if f(x) = ∫[t = x to 0] 1/ (1 + t + t²) dt\n\nf'(x) = -1/ (1 + x + x²)\n\nf''(x) = (1 + 2x)/(1 + x + x²)²\n\nFor concave UP f''(x) > 0\n\nSo 1 + 2x > 0 (as (1 + x + x²)² > 0 for all x)\n\nie x < -1/2
464968
1) If h(x) = f(g(x)), then g(x) = 4+3x and f(x) = x^5\n2) Assuming your function is 3sqrt(x^2 -8), then\ng(x) = x^2 - 8 and f(x) = 3sqrt(x)
416279
You are given P(t=0), Q(t=0); P'(t=0), Q'(t=0); and P''(t=0), Q''(t=0). Given that P'' and Q'' remain the same (that is, the motion of P and Q do not change), you need to show that there is a time T such that P(t=T) = Q(t=T).\n\nIt's late and I'm drunk, so you can do the actual working for yourself :D
120153
(x+ [1+i*sqrt(3)]/2) * (x+[1-i*sqrt(3)]/2) = 0
416483
F ' (x) = 1 / ln x by the fundamental theorem of calculus.\n\n[If you have difficulty understanding why, let g(t) be the antiderivative of f(t) = 1 / ln t. F(x) = g(x) - g(2). Therefore, F ' (x) = g ' (x) = f(x) = 1 / ln x. ]\n\n1 / ln x > 0 for x > 2, which means that F is increasing.\n\nF " (x) = - 1 / (x ln (x) ^2 ), which is negative for x > 2. Therefore, F has negative concavity.
472349
I'll call g(x) = (integral from 0 to x) f(s) ds.\nWe just need to see what g(-x) is. \ng(-x) = (integral from 0 to -x) f(s) ds.\nIf f is even, then f(s) = f(-s), so this equals:\n(integral from 0 to -x) f(-s) ds\nand that is equivalent to\n(integral from 0 to x) f(s) ds = g(x), so g is even.\nSimilarly, if f is odd, f(s) = -f(-s), so it equals\n(integral from 0 to -x) -f(-s) ds\nwhich is\n(integral from 0 to x) -f(s) ds = -g(x), so g is odd.
464650
a)  f(0) = f(0-0) = f(0) - f(0) = 0.\n\nb)  Fix k in the reals, and define f(x) = kx.\n\nc)  Suppose you have f(x)=f(y) for some reals x and y.  Then f(x)-f(y) = 0, and since f preserves subtraction, f(x-y) = 0.  Since 0 is the only element which gets sent to zero, this implies that x-y = 0, and thus x = y.\n\nI have a challenge for you: find a nontrivial function f which preserves subtraction but is not injective.  There are such functions, but they are not necessarily easy to find.  Cheers!
586154
no:\nsqrt(2) + (-sqrt(2)) = 0 rational\nno\nsqrt(2)*(1/sqrt(2)) = 1 rational\n\nsqrt(2) and 1/sqrt(2) are irrational\n\n In general , it's more often that sum ( product) of too irrational is irrational
57941
well, you don't have to worry about sinx and cosx as they are always between -1 and +1. so the first part of the function (1/2)* sinx is bounded.\ncos(x) is bounded.\nSo, you can rewrite the function as,\ny(x) = f(x) - x*g(x).\nwhere both f(x) and g(x) are bounded.\nso, as x goes to infinity, y(x) tends to minus infinity because of -x in the right hand side. \nSimilarly when x goes to minus infinity, y(x) goes to plus infinity\n----------------------\nResponse to the counter-example - \nin case y(x) = 2+x*(-3),\nas x -> infinity y -> -infinity.\nso y(x) is unbounded\nwhat's wrong with that??????
429132
1 is false. The function is continuous.\n2 is false; see above.\n3 is true.  Hence, 4 is false, and so is 5.\nYou are right about the non-differentiability at x=3.  A function cannot be called differentiable if its slope at x- does not approach its slope at x+ as the interval approaches zero.
478127
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...
130019
Consider the function G(x) := 1 + integral (from 0 to x) of f(t) dt.  It is easy to see that G(0) = 1.  It is an easy exercise to show that it is well defined.  And by the fundamental theorem of calculus, it satisfies the last desired condition.
500014
Neither, it is undefined, i.e. it has none of the properties of all rational or whole numbers.\n\nI might be wrong, but I think this is one of those trick questions that are so common in math courses.
29273
If you look at the standard epsilon/delta proof that G(x) = x^2 is continuous 0, it all works for F(x) too. \n\nThat is, given epsilon (e), you can find delta (d) such that /G(x) - G(0)/ < e when /x - 0/ < d.\n(Since for -1 < x < 1,  x^2 < |x|, just pick d = min(e, 1))\n\nNow looking at F(x), given d, pick that same e you would use with G(x). If x is rational, F(x) = G(x)\nso /G(x) - G(0)/ = /F(x) - F(0)/ < e.  If x is irrational, /G(x) - G(0)/ = /0 - 0/ < e.