g(x)={x^2-a^2/x-a} if x does not = a g(x) ={4} if x=a Find the constant a such that the function is continuous
81081
g(x)=(x-a)(x+a)/(x-a)\n= x+a\nSince g(x) =4\ntherefore, x+a = 4 \nbut, x=a so substituting the value of x\na+a=4\n2a=4\na=2\nThe constant a = 2.
586893
If its differentiate x * sqrt 1 + x^2 , the answer is\n  1 + 2x ( since sqrt 1 is a constant and d x^2/dx = 2x)\n\n \nIf your question is sqrt1 + x^2 answer is 0 + 2x = 2x
210475
f(g(x)) = 3/( (3/x)+4) -4)\nso that becomes\n3/ ((3/x)) because the + 4 and the - 4 cancel out\nand when you divide you multiply by recipricol so 3 divided by 3/x = 3 multiplied by x/3\n3 multiplied (x/3) = 3x/3 = x
790936
x≠０\ncuz if x=0, (x-7)/x^3 is meaningless
948694
The anti-derivative of a function f(x) is another function F(x) so that F´(x) = f(x). \n\nIf f(x) = x^n (n<-1 or n>-1) then the anti-derivative is F(x) = (x^(n+1))/(n+1)  + C ... we need to add an arbitrary constant.\n\nIn your case we have:  f(x) = x^1  and  F(x) = (x^2)/2 + C
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
813038
f(x) = x^2 / (x - 2 ) ^2?\n\n    (a) Find the vertical and horizontal asymtotes\nx=2\n    (b) FInd the intervals of increase and decrease\nfind f'(x) = ( (x - 2 ) ^2 2x - x^2 2(x-2) )/(x - 2 ) ^4\n= ( x-2) ( (x-2)2x-2x^2) / (x - 2 ) ^4\n= ( 2x^2 -4x -2x^2 ) / (x - 2 ) ^3\n= -4x/ (x - 2 ) ^3  \n=0\nif -4x =0, i.e. x=0 one critical point.\n    (c) Find the local and maximum and minimum values\n    (d) Find the intervals of concavity and the inflection points
60538
if [ -7x + x^2 =0 ]\nthe you will get y = 7/x
892759
g(t)=1/3(t+2)^2-12 (1<t<4) \nDomain is 1 < t < 4\nRange is -9 < g(t) < 0\n\nSo for inverse\nt = 1/3((g^(-1)(t) + 2)² -12\nSo 3t = ((g^(-1)(t) + 2)² - 36\n((g^(-1)(t) + 2)² = 3t + 36\n(g^(-1)(t) + 2) = √(3t + 36) (NOTE: NOT -√(3t + 36))\n\nSo g^(-1)(t) = √(3t + 36) - 2\nDomain: -9 < t < 0\nRange: 1 < g^(-1)(t) < 4
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
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.
1045860
Function y = f(x) is continuous at point x=a if the following three conditions are satisfied :\n\ni.) f(a) is defined ,\nii.)lim f(x) exists (i.e., is finite) ,\nand\niii.) lim (x ->a)  f(x)  = f(a) \n\nApplying them to your case:\n(1) f(0) is defined, being a^2 and a  in two cases.\n(2)limit also exists here\n(3) f(0)=f(0-)=f(0+)\ntherefore,\na^2=a\nor a(a-1)=0\nor a= 0 or 1.
1082751
1)   f(x) = x+1/x =(x^2+1)/x  \n      g(x) = x-2/x =(x^2-2)/x\n(f+g)(x) = (x^2+x^2+1-2)/x = (2*x^2-1)/x = 2*x -1/x\n\n2)  f(x) = -2/x\n     g(x) = (x+1) / (x-3)\n(f+g)(x) = -2/x + (x+1)/(x-3) = {-2*(x-3)+x*(x+1)}/x*(x-3)\n              = (-2x+6+x^2-x)/(x^2-3x) =(x^2-x+6)/x^2-3x
985294
you just keep on deriving.\n\nI find it easier to split things into exponents.\n\nI am assuming your question goes\n\n(1/(2x))  - 4 = g(x)\nor\n\n(1/2)*x^-1     - 4 = g(x)\n\ndifferentiate once\n\ng'(x) = -(1/2) * x^-2 \ng''(x) = -2 * (-1/2) * x^-3  = x^-3\ng'''(x) = -3 * x^-4  =  -3/(x^-4)  **ans**\n\nPlug 3 into this... bah\n\nFor reference please use parenthesis to make your problem clearer for us. I just solved the whole thing using the wrong g(x).\n\n\n===!using the correct g(x) which should be g(x) =1 / (2x -4)===\n\ng(x) = 1 / (2x -4) = (2x - 4)^-1  don't forget the chain rule\n\ng'(x) = -1*2*(2x - 4)^-2 = -2 * (2x -4)^-2    chain rule again\n\ng''(x) = -2*-2*2*(2x -4)^-3 = 8*(2x -4)^-3    chain rule again\n\ng'''(x) = -3*2*8(2x - 4)^-4 = -48 * (2x -4)^-4\n\ng'''(x) = -48/[(2x -4)^4]\ng'''(3) = -48 * (6 -4)^-4 = -48/ (2^4) = -48/16   = ***real ans***\n\ng'''(3) = -3
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
1112189
1. g(x)=2x-5\nso, g(f(x))=2f(x)-5\nor 2 (x^2+2x+3)-5\nor 2x^2+4x+1\n\n2. \na) put f(x)=y\nso y=2^x+1\n\nnow take independent values of x and find corresponding value of y to have 3 sets of locatins to find the graph\n\nIf x=0, then y=2 (2^0=1)\nif x=1, then y=3\nif x=2, then y=5\n\nb) now you have 3 sets of points (first one stands for x, position them in the graph paper against the function we have just found in (a) and see the domain..
938777
1)  \nTo find increasing and decreasing of a function, you need to take the derivative:\n\nsqrt(4-x) can be re-written as (4-x)^(1/2)\n\nTaking the power rule and chain rule, you get (1/2)(4-x)^(-1/2)(-1)\n\nSimplifying, you get -1/(sqrt(4-x)).\n\nTo determine increasing or decreasing of a function, look at the derivative.  If the derivative is positive, the function is increasing and if the derivative if negative, the function is decreasing.  Looking from the derivative, you can conclude that the derivative is only defined for x < 4 because otherwise, you would have a negative square root which has no real value.  Thus, for all x < 4, you can conclude that the derivative is negative (try plugging in a value for x).  \n\nTherefore, the function is decreasing from (-infinity, 4)\n-------------------------------\n2)\nTake the derivative of the function:\n\nf(x) = x^2 - 4x\nf'(x) = 2x - 4\n\nSetting the derivative equal to zero locates your extreme values:\n\n2x - 4 = 0 ---------------> x = 2\n\nNow perform the number line test to see how the function behaves to the right and left of x = 2.  If you plug in a value to the left of 2, let's say x = 1, you get a negative derivative, thus the function is decreasing.  If you plug in a values to the right of 2, let's say x = 3, the derivative is positive.  Therefore, you can conclude from these findings that x = 2 is a minimum.  Solving for the actual point, plug in x = 2 into original function.  You get y = -4.\n\nSince the function extends to infinity in the y-direction for negative and positive infinty of x, (2, - 4) is an ABSOLUTE MINIMUM\n--------------------\n3)\nAgain, take the derivative:\n\nf(x) = x^3 + 5x^2 -13x + 1\nf'(x) = 3x^2 + 10x - 13\n\nNow set the derivative equal to zero to find extreme values:\n\n3x^2 + 10x - 13 = 0\n(x - 1)(3x + 13) = 0\n\nTherefore, you have extreme points at x = 1, -13/3\n\nNow do number line test like above to see if you get max or min.\n\nRemember, \nnegative->positive change = minimum\npositive->negative change = maximum\n\nx = 1\nto the left (like x = 0) ---------> negative\nto the right (like x = 2) -------> positive\n\nThus, at x = 1, you have a minimum.  The exact point is (1,-6)\n\nx = -13/3\nto the left (like x = -5) ---------> positive\nto the right (like x = -4) -------> negative\n\nThus, at x = -13/3, you have a maximum.  Solve for y to get the exact point.\n\nNow determine values at x = -+ infinity \n\nAt x = +infinity, y = +infinity\nAt x = -infinity, y = -infinity.\n\nTherefore, since at x = 1 and x= -13/3 give you y values that are not the extreme values, these are local extremes.  Thus, x = 1 is a local minimum and x = -13/3 is a local maximum\n-----------\n4)\n\nIf you plug in x = 0, you get a positive value thus it is increasing from -infinity to 1.  if you plug in x = 6, you get a positive value, thus the function is increasing from 5 to infinity.  If you plug in a value between 1 and 5 (like 3), you get a negative value, thus the function is decreasing from x = (1,5)\n--------------\n\nHope this helps