How can Isolve this function: g(x)= Ix-2I+Ix+2I?
114499
Break it down into three sections determined by the points where the absolute values change sign.\n\nfor 2 < x, g(x) = (x-2) + (x+2) = 2x\n\nfor x < -2, g(x) = -(x-2) + (-(x + 2)) = -2x \n\nfor -2 < x < 2, g(x) = -(x-2) + (x+2) = 4
782406
°oO****Oo°°oO****Oo°°oO****Oo°\n\n\n 4 ∫[g(x)] g'(x) dx = 4 x 1/2g(x)² +c = 2g(x)²+c\n\nits the first suggestion that u made but i think you forgot the number 4...\n\n\n°oO~<<_P~'*'_ I~'*'_M~'*'_O~'*'~L _~'*'_A>>~Oo°
772631
Assuming this one is like your other questions, then\nAssuming h(x) = f(g(x)), then \ng(x) = x-1 and f(x) = 2x^5/3 + 5x^2/3
876214
g(x)=-x^3+4x-2\ng'(x)=3x^2+4\ng"(x)=6x\ng"(x)=0=6x\nx=0 is the inflection point
974938
I believe this is how you do it:\n\n(f o g)(x) = f[g(x)] = 3(-x^2) - 4 = -3x^2 -4\n\nNow that this formula is found, sub in 2;\n\n(f o g)(2) = -3(2)^2 - 4 = -3(4) - 4 = -12 - 4 = -16
958895
If you mean y = 2 - x then theabove answers are right.\n\nI'm wondering, though, if you might mean you have a graph of some function f(x) and want to draw a graph of the combined function g(x) = f(2 - x). If so you simply flip the original graph around the y axis (this gives you -x) and then shift the whole thing 2 to the right (this is the 2). You can check individual points if you want to make sure: g(2) = f(0), g(3) = f(-1), ...
917328
Gopal is correct, except (a) he meant x =  -1should be excluded from the domain, and (b) x = 0 should also be excluded from the domain (since g(0) is undefined).
794327
A pretty trivial one would be f(x) = absolute value of x and g(x) = -x\n\nf(g(x)) = abs (-x) = x\n\nbut g(f(x)) = - (abs x) = -x
1057515
Please check the statement of the problem, then apply this.\n\nWe know that\nsin (x)=(e^(ix)- e^(-ix))/2i\n\ncos (x)=(e^(ix)+ e^(-ix))/2\n\nalso\n\nsin(ix)=i(sinh(x))=(e^(-x)- e^(ix))/2i\n\ncos(ix)=(cosh(x))=(e^(-x)- e^(ix))/2\n\nThe only thing you need is if z=x+iy then |z|=sqrt(x^2 + y^2)\n\nGood luck
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.
798580
g=1000[1+(1.3/1200)]^x
764329
x(abs[x- (x^2)] (-1=2)\n\n:> peace\n.
650764
Remember that i^2 is -1 so\n-3i+2(-1)-2+i=-4-2i
781847
(x²+1)/(x+1).\nf(x) = x²+1, g(x) = x+1\n[ 2x*(x+1) - x² * 1 ] / (x+1)²\n(2x²+2x - x²) / (x+1)²\nx²+2x / (x+1)²\n\nf(x) = x²+2x, g(x) = (x+1)²\n(( [ (2x+2)*(x+1)² ] - [ x²+2x * 2(x+1) * 1 ]  )) / ((x+1)²)²\n[ (2x+2)(x²+2x+1) - (x²+2x)(2x+2) ] / (x+1)^4\n[ 2x³+4x²+2x+2x²+4x+2 - (2x³+2x²+4x²+4x) ] / ((x+1)²)²\n[ 2x³+4x²+2x+2x²+4x+2-2x³-2x²-4x²-4x) ] / ((x+1)²)²\n[ 2x+4x+2-4x) ] / ((x+1)²)²\n(2x+2) / ((x+1)²)²\n2(x+1) / (x+1)^4\n2 / (x+1)^3  <----dy²/dx²  Yours was incorrect.\n\nAnd yes, this will never have an inflection point.  It will only approach y=0 as x approaches +/- ∞ and approach x=0 as x approaches 0
906026
Well dear \n\n▪gof(x) = g(f(x)) = g ( √x ) = (√x) - 1 = x - 1\n▪Dgof = { x I x Є  Df , f(x)  Є  Dg }\n▪Df = [ 0 , + ∞ )\n▪Dg = IR\n▪{ x I x Є  [ 0 , + ∞ ) , x Є  IR = [ 0 , + ∞ ) \n\nGood Luck\n{am sure its not your homework , but try to find why the answer would be that }
449247
f(x)=(3x^2)-x+10\ng(x)=1-20x\n\n(f°g)=f[g(x)]\n\nSo we have:\n\nf[g(x)]={[3(1-20x)^2]-(1-20x)+10}\n={3[1-40x+(400x^2)]-1+20x+10}\n=3-120x+(1200x^2)-1+20x+10\n=(1200x^2)-100x+12\n\n(f°g)(5) means that we substitue 5 for x.\n\n[(1200x^2)-100x+12](5)\n={[1200(5)^2]-[100(5)]+12}\n=[1200(25)]-500+12\n=30000-488\n=29512
831528
"What i know: it can also be written as (x^2 + cos x) ^ -1/2\nand the derivitive of that = (2x - sin x) ^ -1/2"\n\nWhat you said here was to take the derivative of the inside function, but what you do is take the derivative of the outside function and multiply it by the derivative of the inside function.\n\nIf you have a composition of two functions such as h(x) = f(g(x)), then h'(x) = f'(g(x)) * g'(x) [or as in your notes, f(g) * g']\n\nThat is why you must use the chain rule for 1 and 3 but not for 2, since 2 does not involve a composition:\n\n1. f(x) = (x^2 + cos x) ^ (-1/2)\n\nf(x) = g(h(x)) where g(x) = x^ -1/2 and h(x) = x^2 + cos x\n\n3.  f(x) = 2 ^ (x^2 - sin x)\n\nThis isn't a product- it's an exponential function of another function.\n\nf(x) = g(h(x)) where g(x) = 2 ^ x and h(x) = x^2 - sin x