Determine   f^(-1) (x), along with any restrictions on its domain, where  f(x) = squareroot(x + 5) , x is lessthan or equal to -4.\n\n\nPlease show all work as to how you got the answer, so I can understand how it works... = ) thanksss!
137040
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.
517895
I think f^(-1) (x) means the inverse of f(x), not the reciprocal.\n\nLey y =f(x)\ny = (x^3 +1)^(1/3)\n\nsubstitute the x and y\nx = (y^3 +1)^(1/3)\n\nnow solve for y\nx^3 = (y^3 +1)\ny^3 =x^3 -1\ny =(x^3 -1)^(1/3)\n\nNow if you put 1.5 into this equation\ny = ((1.5)^3 -1) ^(1/3)\ny =(2.375)^(1/3) or approx. 1.3342
266972
I assume you are asking for the inverse of f(x)\n\nThen f-1(x) = (x+1/3)/2
168758
Here is the problem....\n\nsqrt ((-1)*(-1)) is NOT necessarily equal to sqrt(-1) * sqrt(-1).\n\nEach time you take a square root, you are deciding whether to take the negative root or the positive root - and neg_sqrt(1) is NOT equal to pos_sqrt(1).\n\nrevisit your proof and see what happens when you force sqrt to always equal neg_sqrt throughout the proof or always equal pos_sqrt through the proof.\n\nNeat trap though.\n\nThe simplest way to show the problem is this:\n\n1 = 1\nsqrt(1) = sqrt(1)\n\nso choose positive square root on the left and negative square root on the right, and:\n1 = -1 \n\nta-da!
466101
Just differentiate it. You need f'(x) < 0.\n\nBy the quotient rule, f'(x) = (2x(x^2 + 4) - (x^2)(2x))/(x^2+4)^2.\nNow, since the bottom is a square, thus positive, you just need the numerator to be < 0:\n2x(x^2 + 4) - x^2 * 2x < 0\n2x^3 + 8x - 2x^3 < 0\n8x < 0\nx < 0\nThus it is decreasing on (-∞,0).
41616
the conditions are\n\n(x+2)/(x-2) >= 0\n\nand don't forget\n\nx!=2 (x not equal to 2), even if this latter should come out resolving the first.\n\nFirst inequality can be solved easily; it works as it were a product, save for the fact that you loose the value x=2:\nterms are all with positive x, there are two items in x so it behaves as a parabola with positive a: solution is for values external to the roots: so your domain is\n]-infinity;-2] U ]+2;+infinity[\n\n-2 included, +2 excluded
322089
Domain of the function is the value of x and the range is f(x)....\nfor f(-3) the domain is -3 and the range f(-3)=(-3)^2+2(-3)-48=9-6-48=-45  which is the range....\nsimilarly for f(a+1) the domain is a+1 and the range by substituting a+1 in the equation and factorising you get the range as (a+9)(a-5)...
419975
The function increases where its derivative is positive.\n\nIt decreases where its derivative is negative.\n\nIt has extrema where the derivative is zero.\n\nWhat is the derivative of f(x) = x(x+1)^-1?\n\nI can go further, but it sounds like you're looking for a gentle push instead of a hard shove.
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f(-2) = 5. What's so difficult about this question?
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Note that the function is zero when x=1, this means that you can factor out an (x-1)\n\nx^4-x^2+4x-4 = (x-1)(x^3+x^2+4)\n\nNote that x^3+x^2+4 is zero when x=-2 so you can factor out an (x+2),\n\nx^3+x^2+4 = (x+2)(x^2-x+2)\n\nThe quadratic, x^2-x+2, does not factor with real numbers. (Try out the quadratic equation to confirm.) So the answer is\n\nx^4-x^2+4x-4 = (x-1)(x+2)(x^2-x+2)
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did u mean square root of (2-x) or 1/(2-x)??\ni wil solve both the cases\n\nif it was sqrt(2-x), the solution is like this\nto find out the domain, we need to find all the values of x for which the function is defined. for a square root function, the only requisite is that the term inside the square root should be non-negative.\nso, the condition is (2-x)>=0, ie., x<=2\nso, the domain becomes (-infinity,2]\nnote that parenthsis, ie., '(' or ')' means that the value is not included in the interval, where as square brackets, ie., '[' or ']' means that the value is included.\n\nnow, if the problem is 1/(2-x), the solution is:\nfor this function to exist, the denominator should be non-zero, which means that (2-x) should not be equal to zero.\nso, x should not be equal to 2\nthe domain is given by  (-infinity,infinity) - {2}\nhere, {2} means the set containing the element 2\nthe domain can also be denoted as  R-{2}, where R is the set of all real numbers.
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
539771
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.
396188
x^2 + 3x  = -1\nx^2 + 3x +(3/2)^2 = -1 + (3/2)^2\n(x +(3/2))^2 = (5/4)\nx + 3/2 = sqrt(5)/2 OR x + 3/2 = - sqrt(5)/2\n\n(sqrt means the positive square root) \n\nx = -(3/2) + (1/2)sqrt(5)\nx = -(3/2) - (1/2)sqrt(5)\n\n.......................................................................\n\n2x^2 +4x + 1 = 0\nx^2 + 2x = -(1/2)\nx^2 + 2x + 1 = (1/2)\n(x+1)^2 = (1/2)\n\n(x + 1) = sqrt(1/2) OR (x + 1) = -sqrt(1/2)\n\nx = -1 + sqrt(1/2) OR\nx = -1 - sqrt(1/2)\n....................................\nLW = 783\n\n\n\n(W+2)W = 783\nW^2 + 2W = 783\nW^2 + 2W  + 1 = 784\nW + 1 = sqrt(784)\n\nW = srqt(784) -1\nW + 2 = sqrt(784) + 1\n\n2(W + L) = 2(2) sqrt(784)\n                = 4* 28\n                 =   112
496256
1) sqrt(x - 1) = x - 3\nSquare both sides:\nx - 1 = (x - 3)²\nx - 1 = x² -6x + 9\nx² -7x + 10 = 0\n(x - 2)(x - 5) = 0\nx = 2 or x = 5\n\nAs a double-check:\nsqrt(2 - 1) = 2 - 3\nsqrt(1) = -1 <-- this is a true statement since 1 and -1 are roots.\n\nsqrt(5 - 1) = 5 - 3\nsqrt(4) = 2 <-- check\n\n2) sqrt(y + 12) = 2\nSquare both sides:\ny + 12 = 2²\ny + 12 = 4\ny = -8\n\nAs a double-check:\nsqrt(-8 + 12) = 2\nsqrt(4) = 2 <-- correct\n\nSo your answers are:\n1) x = 2 or x = 5\n2) y = -8
594476
f(x): 'f' is usually a defined function.\n\nf(x) = |4|, would mean the the function returns the absolute value of 'x', which in this case is four.\n\nei: f(-4) = 4. If function is defined as above. written out would be in some programing languages would be f(abs(x)).\n\nf(x) can be any formula you want to define it as.\n\nf(x*y) or f(sin(x))....\n\n\nreally unsure of the how you have stated the problem.
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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.