Given: f(x) is defined everywhere on real line.  f is differentiable everywhere.  \nShow: There exists G(x) such that G(0) = 1 and dG/dx = f(x) for all x.
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.
382781
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
369233
( (x² +1)(y² +1)(z² +1) ) / (xyz) ≥ 8\n\n( (x² +1)(y² +1)(z² +1) ) / (xyz)\n\n= (x + 1/x)(y + 1/y)(x + 1/z)\n\nIf x > 0 least value x + 1/x occurs when x  = 1 (differental calculus, f'(x) = 0 when x = 1 (x > 0))\n\nSo x + 1/x ≥ 2 for all x > 0\nSimilarly y + 1/y ≥ 2 for all y > 0\nand z + 1/z ≥ 2 for all z > 0\n\nTherefore \n(x + 1/x)(y + 1/y)(z + 1/z) ≥ 8 for x, y, z > 0\n\nie ( (x² +1)(y² +1)(z² +1) ) / (xyz) ≥ 8 for x, y, z > 0
556109
nested function\nf(g(x))=f'(x)g'(x)\ng(x) = 7x\nf(x) = ln [7x]\n\nf'(x) = 7\ng'(x) = 1/x\n\nd(ln 7x)/dx = 7/x\n\nNow that I think about it, I'm not sure I like that answer either.  Maybe it's 7/7x or 1/x.
347644
Wait a minute, did you want to know when the function is zero or when its derivative is zero?  \n\nFor derivative:\ny' = 2x + sin x, right?\n\nIt looks like there's one place where it's 0.  Requires and numerical solution.\n\nWell, it's 0 at x=0, right?\nI assume that x is in radians.  1 radian is about 57 deg.  As x decreases from 0 both parts become more negative.\n\nIt looks like x = 0 is the only zero of the dirivative.
440701
df/dx=(df/dg)*(dg/dx)=x’=1, dg/dx=(dg/df)*(df/dx)=x’=1; both are const\nIt means g’(b)=1
501689
f(x) satisfies the mean value theorem because it is continuous and differentiable on [-2, 1].\nSecant slope = (1 - 4) / (1 + 2) = -1\nDerivative f'(x) = 2x\n2x = -1\nx=-1/2 (this is what c equals to)
533035
That uses 'x' a couple of times, so just to make things clearer, I'll call the point we're considering 'y'.\n\nLets call the limits of f and g at y A and B respectively. \nNow, by definition of a limit: \nFor all ε > 0, there exists δ1 > 0 such that for all x with 0<|x-y|<δ1, |f(x)-A|<ε.\nSimilarly,\nFor all ε > 0, there exists δ2 > 0 such that for all x with 0<|x-y|<δ2, |g(x)-B|<ε.\n\nNow, lets suppose (for the sake of contradiction) that B<A. Lets make ε = (A-B)/2. Then there exist δ1 and δ2 such that the above holds.\nNow, from above there exists δ1 > 0 so that if |x-y|<δ1 then |f(x)-A|<ε, and δ2 > 0 so that if |x-y|<δ2 then |g(x)-B|<ε.\n\nLets pick x close enough to y so that both those conditions hold - eg x = y+min(δ1,δ2). Then the following holds:\ng(x) < B+ε = B+(A-B)/2 = A-(A-B)/2 = A-ε < f(x). Contradiction.\n\nThe key idea is that if B>A, then there is a 'gap' between B and A - thus you just need to get smaller than the gap, and you'll get a contradiction.
175505
[ If you haven't covered compactness already, this answer won't be much help. ]\n\n1. The range of this function is equal to its range on the closed interval [0,1].  Since [0,1] is compact, the range is compact.  In particular, it is bounded and closed, so the range has an absolute maximum and minimum.\n\n2. Any continous function on a compact set is uniformly continuous.  So it is uniformly continuous on [0,1], which can easily be seen to be extended to the entire real line.\n\n3. Let g(x) = f(x) - f(x+pi)\n\nThen if x0 is a point where f(x) reaches its maximum, then\n\n   g(x0) = f(x0) - f(x0+pi) >= 0\n\nIf x1 is a point where f(x) reaches its minimum, then\n\n   g(x1) = f(x1) - f(x1+pi) <= 0\n\nSince g is continuous, and it takes one value less than or equal to zero and one value greater than or equal to zero, the intermediate value theorem says g(x') = 0 for some x' between x0 and x1.
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.
501377
For x < 1 \nf(x) = |x-1| + 2\nWhen x → 1-, f(x) →2\nWhen x < 1, |x-1| + 2 = 1 - x + 2 = 3 - x. So slope is -1.\n\nFor x ≥ 1\nFor continuity f(1) = 2.\nFor differentiability f'(1) = -1\n\nSo a + b = 2\n2a + b = -1\n\nThus a = -3, b = 5\n\nf(x) = |x-1| + 2, for x < 1 \n..... = 5x - 3x², for x ≥ 1\nis continuous and differentiable for all x
497325
1) yes, by definition\n2) no, example: sin x on [0,2 pi]\n3) no , example |x|, absolute value of x is continuous in 0  but not differentiable in 0.\n4) yes, since (F-G)' = f-f = 0, then F-G is a constant.\n5) yes, known theorem
494393
f(g(x)) = g(x) + 1 = (x^2 - 1) + 1 = x^2  - 1 + 1 = x^2\n\ng(f(x)) = (f(x))^2 - 1 = ( x + 1 )^2 - 1 = (x^2 + 2x + 1) - 1 \n                                                         = x^2 + 2x + 1 - 1 \n                                                          = x^2 + 2x
433926
Let g(x) = f(x) - f'(x). Then the problem can be restated as saying that g(x) has at least one real root. If f(x) is a polynomial of degree n, where n≠0, then f'(x) is a polynomial of degree n-1, and so the difference of the two polynomials will be a polynomial of degree n. We consider two cases:\n\nSuppose n is odd. Then g(x) has odd degree, and all polynomials of odd degree have a real root, so we are done.\n\nSuppose n is even. Then f(x) has either an absolute minimum or an absolute maximum. Suppose without loss of generality that it has an absolute maximum at c (the arguments for polynomials with an absolute minimum are essentially the same). Then at that point f'(c) = 0, so g(c) = f(c). Since f(x) has a real root, f(c) ≥ 0. If f(c) = 0 then g(c) = 0 and we are done. Otherwise, consider that the x^n term of f(x) dominates the terms of f'(x), so [x→∞]lim g(x) = -∞. Thus, g(x) is negative at at least one point, and since g(x) is everywhere continuous, then by the intermediate value theorem it must have at least one zero. Q.E.D.\n\nNote that in the special case where n=0, f(x) is a constant function, and thus does not have a root, so it does not meet the conditions of the theorem. Finally, in the case where f(x) = 0 (which is usually considered to be a polynomial of indeterminate degree, rather than degree zero, primarily to avoid exceptions to the fundamental theorem of algebra), f(x) is everywhere equal to f'(x).
489996
You didn't write your problem clearly.  I don't know if g(x) is:\n\n1)  x² + (7/x³) - 2, or\n2)  (x² + 7) / (x³ - 2)\n\nThe quality of the answers you receive are in direct relation to the quality of the question you write!  :-)\n\nAssuming you meant (1):\n\ng '(x) = 2x - 21x^-4\n\nAssuming you meant (2):\n\ng '(x) = (-x^4 - 21x² - 4x) / (x³ - 2)²\n\nEDITED TO ADD:\n\nHi Jeff, I got your email but couldn't respond; yahoo said your email was "unconfirmed" whatever that means!  So, I am responding to you here. Hope you see this!\n\nYOU WROTE: "I was wondering if you could explain how to work through this problem"\n\nMY RESPONSE:\n\nHi Jeff, sure I am happy to explain. Basically, when you differentiate a fraction f(x) over g(x) the rule is "bottom times derivative of top, minus top times derivative of bottom, all over bottom squared" -- i.e., \n\n[ (g(x)*f '(x)) - (f(x)*g '(x)) ] / (g(x))^2\n\nSo in your problem f(x) = x^2 - 7\ng(x) = x^3 - 2\n\nSo derivative = [ (x^3 - 2)*(2x) - (x^2 - 7)*3x^2 ] / (x^3 - 2)^2\n\nThen just multiply the numerator and combine like terms to get:\n\n[ -x^4 - 21x^2 - 4x ] / (x^3 - 2)^2\n\nDid that help?  Let me know if you have additional questions!\n\n.
506378
if f is differentiable in x+0 then its also continue in x_0 :\n\nf(x) - f(x_0) = f'(x_0)(x-x_0)+ E(x)\n\nlim x->x_0 then left side 0 and right side 0 thus f(x) = f(x_0) for x->x_0\nthus continous in x_0
393069
1. f(g(x))\n2. g(f(x))