Specify the domain and image set of g^-1 (i.e the inverse function of g) and find its rule.  \n\nI have tried to do this and I get g^-1(x)=[sqrt(2x+36)] + 3 where x can be a minimum of -16 and maximum of 9 as the inverse function.\n\nThe domain I get in interval notation is [-16,9] and image of [5,9].\n\nI am confused about the image set and the domain for the inverse function of g. What they mean and what they are and whether my answer is right or not.\n\nCan anyone who is good at maths please see if they can solve this one and explain the answer in an easy and understandable way.\n\nThank you.
122338
A good start for the range and image of an inverse function is that they are in general the reverse of the original function.  The range of the inverse function is the image of the original function and the image of the inverse function is the range of the original function.
402109
To have an inverse function, the original function must be one-to-one. i.e. There must be only one value of x to any value of y in the range.  e.g. y = x^2 (all x) hasn't an inverse function because you can get y = 9 from both x = 3 and x = -3.  If you restrict x to non-negative values, then there is an inverse -- the positive square root.\n\nIn fact, there is no such problem for (x-5)^(-1), because it's already one-to-one.  However it isn't defined for x = 5  (division by 0) and so any interval which doesn't include 5 is acceptable.\n\nMake x the subject of the equation\ny = (x-5)^(-1)  \n(you should get x = 1/y + 5)\n\nthen swap x and y:\ny = (1/x) + 5, and so this is your inverse function.\n\nThe second example is a bit harder.  Think graphically.  The axis of symmetry of the parabola is\nx = -1,   [x = -b/(2a)  = -4/(2*2)]\n\nso if you restrict x to values which which are all on one side of this axis -- left or right, it doesn't matter -- you'll have a one-to-one function.  However, solving\ny = 2x^2 - 7 + 4x\ninvolves using the quadratic formula on\n\n2x^2 + 4x - (y + 7) = 0 and I'm going to be a bit lazy and leave you to finish that.  Any problems email h_chalker@yahoo.com.au
43845
f(x) = x^1/2\ng(x) = x+3\n\nFoG(x)= (x+3)^1/2\nDomain: Since you can't take the sq root of a negative number, X can only be -3 to positive infinity.\n\nGoF(x)= x^1/2 +3\nDomain: X can only be 0 to infinity\n\nCheck the question and answer again.  It may be asking for GoF(x)
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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??????
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Say the point (1,2) is on f(x) and the point (2,5) is on g(x).\nNow since (1,2) is on f that means f(1) = 2.\nSimilarly, g(2) = 5 since (2,5) is on g.\nYou can now find (gof))(1)\n(gof)(1) = g[f(1)]  = g{2) = 5\n(1,5) is a point on (gof)(x)\nThe y-value of one function has to match the x-value of the other one.\nLet (4,7) be on f and (3,4) be on g.  You can find (fog)(3)\n(fog)(3) = f(g(3)) = f(4) = 7\n(3,7) is a point on (fog)(x)
615291
Let's look at the things that can cause trouble with a function as regards its' domain.  Notice that the expression inside of the radical must be non-negative, since we aren't allowed to take the square root of a negative number (in this context.)  So, we must have that x^2+1>=0.  This statement is true for all real numbers x, since for any real number x, x^2>=0, so x^2+1>=1>0.  Also, the expression in the denominator, sqrt(x^2+1), cannot equal zero.  This can't happen, though, because we already know that x^2+1>0 from our work above.  Thus, the domain of the function is the set of all real numbers.
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I think the problem you meant is:\n\n ((2X+1)^(1/3))/(X²-10X-11).  \n\nThe domain (for real numbers) is limited by two things:  Division by zero, and the even root of a negative number.\n\nHere the denominator can be factored into (X-11)(X+10) and that will equal zero at +11 and -10.  The cube root of a negative number is not a problem.  So the domain is all real numbers not equal to +11 or -10.
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The domain is [0, pi/4), (pi/4, 3pi/4), (3pi/4, 2pi]\n\nThe range is [-1, -infinity), [1, infinity).\n\nJust in case, [ means "number next to it included" and ( means "number next to it not included." so [a, b) means from a (a included) to b (b not included.)
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The domain for a function is all the "x" values that work in the functions.\n\nAn example is "y = x"\nIn this case you can plug anything in for "x" and get a "y"\n\nAnother example is "y = 1/x"\nHere, x cannot be zero (since you cannot have 1/0)\nTherefore the domain is all values except x = 0\n\nIn your equation, (1/x^2)  - 3x  + 2, the domain is all real numbers except x = 0.  If x were 0, you'd have (1/0) - 0 + 0 and you cannot evaluate that.
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composition means like g(f(x))?\n\ndomain or result is a sub-set of domain of inner function (f),\nrange is a subset of the range of outer function (g)
137257
sec is 1/cos. So the domain is the set of all real numbers except when cos(2x - pi) is 0. To graph it, first plot the cosine graph corresponding to -2cos(2x - pi). Then, "flip" the wave along y=2 and y=-2 for when the graph is positive and negative. Basically, what you get are sets of parabolas going off to infinity alternating between being above and below the x axes. Then, erase your cosine graph and shift these "parabolas" up by 3, and you're good.\nSince -2cos can take any value between -2 and 2, -2sec can take any value except those between -2 and 2. Shift up by 3, and you get a range of all real numbers less than or equal to 1 and greater than or equal to 4.
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The inverse function is computed as below. In general, if you have a function of the form,\n\nf(x) = y, the inverse function is defined as \nx = f^(-1) (y), basically rearrange the equation so that you have the x on the left hand side of the equation.  This may not be always possible.\n\nIn the present case, when you solve for x,\ny = sqrt(x+5)\ny^2 = x + 5\nx = y^2 - 5\n\nhence the inverse function is y^2 - 5.
36512
OK, first off no one seemed to be able to answer the dividing by 0 and 1 issue correctly, so here goes...\n\nDivision by n is defined as multiplication by the multiplicative inverese of n.  The multiplicative inverse of n is defined as the number 1/n such that n*(1/n)=1.  Zero has no multiplicative inverse because 0 times anything is 0.  Therefore division by zero implies multiplication by a number that does not exist.  One does have a multiplicative inverse (1x1 = 1), so division by 1 is perfectly valid, although you are correct that the result is the number which was divided by 1.  It is exactly the same as multiplying by 1.\n\nNow on the area of a circle issue...\n\nYou seem to be trying to relate area to circumferance.  It's not hard.\nLet A = the area of the cirlce, c = the circumferance, and r = the radius.\nYou already know that A=(pi)r^2 and c=2(pi)r.\nSolve the second equation for r and get r=c/(2pi).\nNow substitute that into the first equation to get A=(pi)(c/(2pi))^2.\nThen simplify and you get A=(c^2)/(4pi).\n\nLike it or not, you can't get rid of the pi.  Pi is defined as the ratio of the diameter of a circle to its circumference, so it's going to be there no matter what.  When you say it's useless unless you know all the digits, this simply isn't true.  Depending on what you're doing, you can either get the pi's to cancel out, or you will be finding a decimal approximation of the area or circumference.  If the radius is a whole number, the area simply won't be (there may be a few values for which it will be, but most of the time it isn't), and any formula you come up with that says otherwise is incorrect.\n\nAs for the sphere question, I don't understand what you're asking.  Are you trying to equate area to volume?  They're not the same thing.  You can't find the area by sticking together infinitely thin slices as the slices have no height so you will never have a third dimension.  You might be able to do it using limits, but that's not going to be the kind of nice simple numbers you're looking for.  If you can explain what you meant, I might be able to do a bit better.
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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.
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The Edifier has it right, but darned if I can follow the notation. You really need to see this written out like a textbook to understand what's going on.\n\nClick the link below in "sources" to see the answer hosted at my website www.calculus-help.com. Right now, Yahoo! answers, as far as I can tell, doesn't allow graphics, which makes it tricky to answer ANY math questions.\n\nPlease don't post math answers, guys, if the poster knows the answer but can't get there, and you get something completely different.
382253
the domain and the range of h(x):\nthe domain : x is less than or equal to 2,the range : y is greater than or equal to zero.\n\nDomain and range for the inverse of h(x) :\nD: x is grater than or equal to zero.\nR: y is less than or equal to 2.\nNotice that the range of h(x) is the same as the domain of   the inverse  of h(x).And the domain of h(x)is the same as the range of the inverse of h(x).
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You differentiated and found it has relative min at x=1/3?  but if you differentiate again you find second deriv is\n54x - 1/(x^2), and I think this goes from \nnegative when x<1/cuberoot(54) to positive when it's greater, so that means it has an inflection.  Also means it's concave down when \nx<1/cuberoot(54), so the answer must be 5, not 2.  Unless I made a mistake!\n\nPS  Dave is wrong about 1.  Domain is all positive real numbers, because lnx is undefined for 0 and neg.