If Lg(x)=f(x) is known and g(x) wanted to be found.How can one propose that :?
603036
The idea is that the y_n form an orthonormal basis so every function can be expanded as a series of the form \nsum a_n y_n.\n\nHow is this known? Typically, the inverse operator of L can be shown to be a compact operator and the general theory of compact operators can be invoked. Some care may be needed with the convergence properties of the series you obtain. Only mean-square convergence (L_2) is guaranteed. \n\nIn any case, the fact that the y_n are eigenfunctions for L and that L is linear makes the determination of the A_n from the B_n much easier. The convergence details can either be ignored or dealt with separately.
776562
f(g(x)) = 3(2x-7)+5\n= 6x-21+5\n=6x-16\n\ng(f(x)) = 2(3x+5)-7\n=6x+10-7\n=6x+3
893232
f(g(x)) = f(√5 * x)\n= cot(π/10 * (√5 * x)\n= cot(πx/(2√5))\nWhen x = 1\nf(g(x)) = cot(π/(2√5))\n≈ 1.1813
819082
Let f(x) = x ³/ (x² + 2x - 15) \n\n= x - 2 + (19x -30)/(x² + 2x - 15) (by division)\n\n= x - 2 + (19x -30)/[(x - 3)(x + 5)]\n\nSo f(x) is undefined when x = -5, 3\n\nAlso lim x→±∞ f(x) = x - 2\n\nThus f(x) is asymptotic to g(x) = x - 2 as x →±∞ and asymptotic to x = -5 and x = 3
935264
Actually, the answer is *not* 4+x.\n(f/g)(x) = (16-x^2)/(4-x)\nNow, factorise the top:\n(4+x)(4-x)/(4-x).\nSo *if x is not 4* then (f/g)(x) = 4+x. However, the function is not defined at x = 4.
30225
These problems are talking about composite functions, which is where the results of one function are input into the second.  Usually it's written with a small circle in between (I'll use gof(x)) and read "g of f of x".  Remember that the function closest to the variable is the one that acts on it first.\n\nOne helpful tool to use when you can't figure out a problem is simplifying it to a problem you've done before.  For example, if f(x)= 3x and g(x)=x^2 ("x squared") then the composition gof(x) means take the results of f(x) and put them into g(x).  \nSo then gof(x) = (3x)^2 = 9x^2.\n\nYour problem is a bit more complicated, but we can use the same process:  \n\ngof(x) would be g(ax-1) so \ngof(x) = (ax-1)^2 + 3(ax-1) + 1\nUse the distributive property to expand this to\ngof(x) = a^2x^2 - 2ax + 1 + 3ax - 3 + 1\nand combine like terms to get\ngof(x) = a^2x^2 + ax - 1   *\n\nSo far, we've done everything just from the definitions of g(x) and f(x).  Now we use the last piece of information we were given: \nthat gof(x) = 4x^2 + ax + b  *\nWe compare the coefficients on like terms of these two polynomials (with the stars) to determine that since a^2 must equal 4, \na = 2 and b = -1 (from comparison of the constant terms).\n\n\nSee if you can use a similar process for the second problem!
575758
In the first problem, where you know g(x) = x + 4; and \nf(x) = 10x;  if you want to find g(f(x)), that means in the expression g(x), substitute 10x where ever you see x.  If you want to figure out f(g(x)), you would substitute (x + 4) for x in the expression 10x, or you would have 10(x + 4) = 10x + 40.\n\nI am going to assume for the second problems that the exponents are -5 and 3/4 respectively.\n\nf(g(x)) = 4(x^(3/4))^-5 = 4x^(-15/4)\n\ng(f(x)) = (4x^-5)^3/4 = 4^(3/4) x^(-15/4)
972739
f(x)=-5(x)+7\nf0g=-5[-9x^2-4x+5]+7\n     =45x^2+20x-25+7\n     =45x^2+20x-18\nfog(-9)\n=45(-9)^2+20(-9)-18\n=3645-180-18\n=3447
965701
f(x) = x(x+2)(x-2)(x-4)\n\nMultiply it out if you need it in general form.\n\nThere could also be a constant in front of the x.
865303
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)
155864
yes, it equals x(ln(x)-1)
882612
2x^2 ln(x) - x^2\n\nVerify by differentiating\n2x + 4x ln(x) - 2x = 4x ln(x)
738204
Wouldnt that just be Work=Work\n=> m x g x h = m x (lambda)\n=> gh=(lam) ?
818927
if you want f(3)\nyou will have f (2 +1) which leads us to know that 2 is x...\nsubstitute all x's with 2's and then your answer will be revealed
795785
f(x): 1/x-3\ng(x): x^2\n\nf(g(x)) means taking g(x) and plugging it into f(x).  So, we need to find an f(x) and g(x) that produce the fog(x) when you plug g(x) into f(x).  Let's pick an easy one for f:\n\nf(x) = 1/x-3\n\nBut h(x) = f(g(x)), so\nf(g(x))=1/g(x)-3\n\nWhat must g(x) be?  From what we know of h(x), it has to be x^2\n\n\nHope this helps.
910126
I'm assuming this is with metric spaces D with the metric d and ℝ with the Euclidean metric.\n\nProof.\n\nLet ε > 0. Since both f and g are uniformly continuous, ∃ δ₁, δ₂ > 0 such that for ∀ x, y ∈ D,\n\nd(x, y) < δ₁ ⇒ |f(x) - f(y)| < ε / 2, and\nd(x, y) < δ₂ ⇒ |g(x) - g(y)| < ε / 2.\n\nLet δ = min(δ₁, δ₂). Then d(x, y) < δ implies\n\n|(f+g)(x) - (f+g)(y)|\n= |f(x) + g(x) - f(y) + g(y)|\n= |f(x) - f(y) + g(x) - g(y)|\n≤ |f(x) - f(y)| + |g(x) - g(y)|\n< ε / 2 + ε / 2\n= ε. ∎
566883
First, since 1/x is not defined for x = 0, before you can show  anything about your functions, you have to define values for g(0) and f(0). I'm going to assume that either the original statement of the problem had that or that you are supposed to find the value that makes the problem solvable.\n\nTo be true you have to define the problems with f(x) = 0 and g(x) = 0\n\n\nFor the continutiy of f(x) given epsilon > 0 find delta > 0 such that |f(x) - f(0)| < epsilon for all |x-0| < delta.\n\n|f(x) - f(0)| = |xsin(1/x) - 0| < |x| = |x - 0|\n\nGiven epsilon, pick delta = epsilon.\nIf |x-0| < delta, then |f(x) - f(0)| < epsilon.\n\n\nAs for the differentiability of g(x),\n\nf(0+h) - f(0)) = (0+h)^2sin(1/(0+h)) - 0\n\nlim(x->0)(f(0+h) - f(0))/h = \nlim(x->0)((0+h)^2sin(1/(0+h)) - 0)/h = \nlim(x->0) h^2(sin(1/h)/h = lim hsin(1/h) = 0\n\nSo the limit exists and is 0.